• —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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, right: '

  • —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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We use cookies to ensure a smooth browsing experience. By continuing we assume you accept the use of cookies.
Learn More
, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;If the probability of getting three consecutive heads or three consecutive tails when tossing 15 coins can be written as $\\frac{a}{b}$, and $a,b$ are positive and coprime. 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It does not work.\u003c\/span\u003e\u003c\/p\u003e\u003cp class=\" s-rich-text-wrapper\" style=\"text-align: center; font-size: 17px;\"\u003e\u003cspan style=\"color: #50555c;\"\u003eThe answer must be an integer from -1 googol to 1 googol(10^100).\u003c\/span\u003e\u003c\/p\u003e\u003cp class=\" s-rich-text-wrapper\"\u003e\u003cspan style=\"color: #50555c;\"\u003eOur problem and the answer of last week problem is released every Saturday.\u003c\/span\u003e\u003c\/p\u003e\u003c\/div\u003e","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":null,"lastLineTextAlign":null},"defaultDataProcessed":true}}}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"Home","uid":"88bf8fa0-779b-4b60-b916-8016e4ce4239","path":"\/home","memberOnly":null,"paidMemberOnly":null,"buySpecificProductList":null,"specificTierList":null,"autoPath":true,"authorized":true},{"type":"Page","id":"f_3536d300-ca00-49e6-b141-407d00199cf3","sections":[{"type":"Slide","id":"f_d9fc6e76-290e-4dad-ac0e-a025dacb393c","defaultValue":null,"template_thumbnail_height":"184.02681333333334","template_id":null,"template_name":"hero","template_version":"s6","origin_id":"f_ed49430a-6806-4c86-b5ca-4989dff8dad4","components":{"slideSettings":{"type":"SlideSettings","id":"f_0b295b32-1ff8-42a2-a190-e262778e5794","defaultValue":null,"show_nav":false,"show_nav_multi_mode":false,"nameChanged":true,"hidden_section":false,"name":"CONTACT","sync_key":null,"layout_variation":"images-right","padding":{"top":"large","bottom":"large"},"layout_config":{"width":"auto","height":"auto","vertical_alignment":"middle","large_legacy":true}},"background1":{"type":"Background","id":"f_c5c8d106-a293-41ac-8a15-9ee2f380fd85","defaultValue":true,"url":"","textColor":"","backgroundVariation":"","sizing":"","userClassName":"","videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":false,"focus":null,"backgroundColor":{"themeColorRangeIndex":null,"value":"#FFFFFF","type":null}},"text1":{"type":"RichText","id":"f_0d05c99f-47ef-4249-93dd-2aa0606994e9","defaultValue":true,"value":"\u003cdiv class=\"s-rich-text-wrapper\" style=\"display: block;\"\u003e\u003cp class=\"s-rich-text-wrapper font-size-tag-custom\" style=\"font-size: 100px;\"\u003e\u003cspan class=\"s-text-color-default\"\u003eA Heroic Section\u003c\/span\u003e\u003c\/p\u003e\u003c\/div\u003e","backupValue":null,"version":1},"text2":{"type":"RichText","id":"f_a81e0fb3-a16a-413c-8d6c-79dd9fa40fea","defaultValue":true,"value":"Introduce your product or service!","backupValue":null,"version":null},"block1":{"type":"BlockComponent","id":"f_064d4986-b0eb-4b62-b303-badbb7c163d1","defaultValue":true,"items":[{"type":"BlockComponentItem","id":"f_924cde25-ebf6-4842-a6fb-8d0e75387d2d","defaultValue":true,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_53fbd497-bdc7-4fd2-b1d0-9cf764d30005","defaultValue":true,"items":[{"type":"BlockComponentItem","id":"1d16a946-549c-4c6a-818a-d8cc95f47d86","defaultValue":true,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"07ebb51d-1d2c-4e21-9be5-3517fad69843","defaultValue":true,"items":[{"type":"Media","id":"f_f05cde53-b4ac-475a-b170-b3cfa0a3bde0","defaultValue":true,"video":{"type":"Video","id":"f_641fa8d5-02e5-4059-a633-9aa59328edf1","defaultValue":true,"html":"","url":"https:\/\/vimeo.com\/18150336","thumbnail_url":null,"maxwidth":700,"description":null},"image":{"type":"Image","id":"f_29839431-60d7-4a63-b56c-5f9c7538cf3d","defaultValue":true,"link_url":"","thumb_url":"!","url":"!","caption":"","description":"","storageKey":"7258853\/401255_206963","storage":"s","storagePrefix":null,"format":"png","border_radius":null,"aspect_ratio":"1\/1","h":800,"w":1200,"s":4082960,"new_target":true,"focus":null},"current":"image"}]}}},{"type":"BlockComponentItem","id":"f_1069b9c8-d910-4124-a9e0-7da1b492efc9","defaultValue":true,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_6e65ccb6-618e-41f4-b121-168a396c93a2","defaultValue":true,"items":[{"type":"BlockComponentItem","id":"f_555f1b40-c560-457c-9b54-bb08607e1bf6","defaultValue":true,"name":"context","components":{"text1":{"type":"RichText","id":"f_c5c042e7-8fa0-445a-86e2-362158930a9e","defaultValue":false,"value":"\u003cdiv class=\"s-rich-text-wrapper\" style=\"display: block;\"\u003e\u003cp class=\"s-rich-text-wrapper s-rich-text-wrapper s-rich-text-wrapper s-rich-text-wrapper\" style=\"text-align: left; font-size: 18px;\"\u003e\u003cspan style=\"color: #000000;\"\u003e\u2014\u2014 Contact\u003c\/span\u003e\u003c\/p\u003e\u003c\/div\u003e","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":"left","lastLineTextAlign":"left"}}}},{"type":"BlockComponentItem","id":"f_687487b6-32ec-4988-bd9e-90104bc357f5","defaultValue":true,"name":"title","components":{"text1":{"type":"RichText","id":"f_f5c90e32-8d63-43b0-a33e-e7df45c6cfc5","defaultValue":false,"value":"\u003cdiv class=\"s-rich-text-wrapper\" style=\"display: block; \"\u003e\u003ch3 class=\" public-DraftStyleDefault-block public-DraftStyleDefault-ltr s-title s-font-title s-rich-text-wrapper font-size-tag-header-three s-text-font-size-over-default\" style=\"text-align: left; font-size: 24px;\"\u003e\u003cspan style=\"color: #1d2023;\"\u003eInterested in the project? \u003c\/span\u003e\u003c\/h3\u003e\u003ch3 class=\" public-DraftStyleDefault-block public-DraftStyleDefault-ltr s-title s-font-title s-rich-text-wrapper font-size-tag-header-three s-text-font-size-over-default\" style=\"text-align: left; 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font-size: 48px;\"\u003eWhat do we do?\u003c\/h1\u003e\u003c\/div\u003e","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":"left","lastLineTextAlign":"left"}}}}]}}}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"f_258a12ae-ad81-4133-bb8e-c5d723bdbf0a","defaultValue":true,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"eb9ec456-1879-4acb-a036-017272f73dda","defaultValue":true,"items":[{"type":"BlockComponentItem","id":"36292aa5-545e-40fd-af4d-971f42d89379","defaultValue":true,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_afa964ba-fd62-46a8-847f-e6109d8338c3","defaultValue":true,"items":[{"type":"BlockComponentItem","id":"f_d802d2db-d9ee-4c54-86b1-b161a8bc8991","defaultValue":false,"name":"smallFeatureLeft","components":{"media1":{"type":"Media","id":"f_581c34bc-1312-4168-ab35-b5bd52bcb6fe","defaultValue":true,"video":{"type":"Video","id":"f_1797ecac-e6a1-4ed5-b4e4-4164153c34a5","defaultValue":true,"html":"","url":"https:\/\/vimeo.com\/18150336","thumbnail_url":null,"maxwidth":700,"description":null},"image":{"type":"Image","id":"f_abfd5e2e-ed3a-4a9c-92e3-a29825acf820","defaultValue":true,"link_url":"","thumb_url":"https:\/\/user-images.strikinglycdn.com\/res\/hrscywv4p\/image\/upload\/f_auto,w_200,c_max,q_auto\/unsplashcom\/photo-1614621972350-8e78b7f5269d","url":"https:\/\/user-images.strikinglycdn.com\/res\/hrscywv4p\/image\/upload\/f_auto,q_auto,w_4096\/unsplashcom\/photo-1614621972350-8e78b7f5269d","caption":"","description":"","storageKey":"https:\/\/user-images.strikinglycdn.com\/res\/hrscywv4p\/image\/upload\/f_auto,q_auto,w_4096\/unsplashcom\/photo-1614621972350-8e78b7f5269d","storage":"un","storagePrefix":null,"format":null,"border_radius":null,"aspect_ratio":null,"h":810,"w":1080,"s":null,"new_target":true,"focus":null},"current":"image"},"text1":{"type":"RichText","id":"f_13ec13e5-05a1-4412-86d2-f2ccb164600f","defaultValue":false,"value":"","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":null,"lastLineTextAlign":null},"defaultDataProcessed":true},"text2":{"type":"RichText","id":"f_77f51541-83b4-4a97-bfdd-0022ebd1c559","defaultValue":false,"value":"\u003cdiv class=\"s-rich-text-wrapper\" style=\"display: block;\"\u003e\u003cp class=\" s-text-color-default s-rich-text-wrapper\" style=\"text-align: left; font-size: 18px;\"\u003e\u003cspan style=\"color: #000000;\"\u003e\u003cstrong\u003eOne Problem a Week (OPAW)\u003c\/strong\u003e\u003c\/span\u003e\u003cspan style=\"color: #000000;\"\u003e is an independent math initiative dedicated to cultivating deep thinking, curiosity, and problem-solving skills.\u003c\/span\u003e\u003c\/p\u003e\u003ch4 class=\" s-text-color-default s-rich-text-wrapper font-size-tag-header-four s-text-font-size-over-default\" style=\"text-align: left; font-size: 20px;\"\u003e\u003cspan style=\"color: #000000;\"\u003eEach week, we publish one carefully designed original mathematics problem. The goal is not speed or competition, but clarity of thought, creativity, and persistence. We believe that meaningful progress in mathematics comes from engaging deeply with a single problem, rather than rushing through many.\u003c\/span\u003e\u003c\/h4\u003e\u003ch4 class=\" font-size-tag-header-four s-text-font-size-over-default\" style=\"text-align: left; font-size: 20px;\"\u003e\u003cspan style=\"color: #000000;\"\u003eParticipants are encouraged to explore different approaches, share partial ideas, and reflect on their reasoning. Complete solutions are released the following week to allow time for genuine thinking and discussion.\u003c\/span\u003e\u003c\/h4\u003e\u003cp class=\" s-text-color-default s-rich-text-wrapper\" style=\"text-align: left; font-size: 18px;\"\u003e\u003cspan style=\"color: #000000;\"\u003eOPAW is open to students and math enthusiasts of all backgrounds who enjoy challenging problems and thoughtful exploration\u003c\/span\u003e\u003c\/p\u003e\u003c\/div\u003e","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":"left","lastLineTextAlign":"left"},"defaultDataProcessed":true},"button1":{"type":"Button","id":"f_8c8ee777-7860-4a13-80ec-aff97d5c70bd","defaultValue":false,"alignment":"left","text":"","link_type":"Web","size":"medium","mobile_size":"automatic","style":"","color":"","font":"Montserrat","url":"","new_target":false,"version":"2"}},"layout_config":{"noTemplateDiff":true,"card_radius":"medium","border_thickness":"medium","card_padding":"medium","spacing":"S","subtitleReplaceToText":true,"mediaSize":"xs","horizontal_alignment":"left","layout":"E","border":false,"mediaPosition":"left","showButton":true,"border_color":"#cccccc","card":false,"structure":"list","card_color":"#ffffff","vertical_alignment":"middle","grid_media_position":"top","columns":"four","isNewFeatureList":true,"content_align":"center"}}],"inlineLayout":null}}}],"inlineLayout":"12"}}}],"inlineLayout":"1"}}}],"title":"About us","uid":"f35a3dd5-e4c5-4cdd-a63a-49ed125fecf8","path":"\/about-us","memberOnly":null,"paidMemberOnly":null,"buySpecificProductList":null,"specificTierList":null,"autoPath":true,"authorized":true},{"type":"Page","id":"f_e666383d-9571-4938-af21-ac778decd2db","sections":[],"title":"History","uid":"42449f65-a5f7-4277-bae9-5af5cab068e9","path":"\/history","memberOnly":true,"paidMemberOnly":null,"buySpecificProductList":{},"specificTierList":{},"autoPath":true,"authorized":true},{"type":"Page","id":"f_3f302fdf-8163-495e-9b73-8c405d9df7f6","sections":[{"type":"Slide","id":"f_ca5cbbd9-b549-4809-9610-51d4a75e1028","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_87b4c413-3185-4681-b539-7a66dc73415f","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_b8f2dd0a-be56-4e46-84d7-01583ad06533","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_5a1f82b2-37f3-466f-9920-e90ba55997f4","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"c997ca39-e805-477b-a756-43a2eecf449d","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"e8de812f-247e-49d4-ab08-f9a21af078df","items":[{"type":"BlockComponentItem","id":"45898861-04cd-4e11-b192-37db18b8a106","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"82c8653b-06c6-498d-bd0c-f381f647e120","items":[{"type":"HtmlComponent","id":26526664,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem\u0026lt;\/title\u0026gt;\n \u0026lt;!-- Load MathJax --\u0026gt;\n \u0026lt;script src=\"https:\/\/polyfill.io\/v3\/polyfill.min.js?features=es6\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script id=\"MathJax-script\" async\n src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-mml-chtml.js\"\u0026gt;\n \u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nGiven that\n\\[\nx^4 - 4x^3 + 6x^2 - 4x = 2022,\n\\]\nlet \\(P\\) equal the product of all non-real roots. Suppose \\(P\\) can be written as\n\\[\nP = a + \\sqrt{b}, \\quad \\text{where } a, b \\in \\mathbb{Z}.\n\\]\nFind the number of factors of\n\\[\n2025b^{2026}.\n\\]\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;p\u0026gt;\n\nWe are asked to solve\n\\[\nx^4 - 4x^3 + 6x^2 - 4x = 2022\n\\]\nand find the number of factors of \\(2025b^{2026}\\), where \\(b\\) comes from the product of non-real roots.\n\n\nNotice that\n\\[\nx^4 - 4x^3 + 6x^2 - 4x + 1 = (x-1)^4.\n\\]\nThus, we can rewrite the given equation as\n\\[\nx^4 - 4x^3 + 6x^2 - 4x = 2022 \\implies (x-1)^4 - 1 = 2022 \\implies (x-1)^4 = 2023.\n\\]\n\n\nTaking fourth roots (including complex roots), we have\n\\[\nx - 1 = \\pm \\sqrt[4]{2023}, \\quad \\pm i \\sqrt[4]{2023}.\n\\]\nTherefore, the roots are\n\\[\n\\begin{aligned}\nx_1 \u0026amp;= 1 + \\sqrt[4]{2023}, \\\\\nx_2 \u0026amp;= 1 - \\sqrt[4]{2023}, \\\\\nx_3 \u0026amp;= 1 + i \\sqrt[4]{2023}, \\\\\nx_4 \u0026amp;= 1 - i \\sqrt[4]{2023}.\n\\end{aligned}\n\\]\n\n\nThe non-real roots are \\(x_3\\) and \\(x_4\\):\n\\[\nx_3 = 1 + i \\sqrt[4]{2023}, \\quad x_4 = 1 - i \\sqrt[4]{2023}.\n\\]\nTheir product is\n\\[\nx_3 x_4 = (1 + i \\sqrt[4]{2023})(1 - i \\sqrt[4]{2023}) = 1 + (\\sqrt[4]{2023})^2 = 1 + \\sqrt{2023}.\n\\]\nThus, in the form \\(P = a + \\sqrt{b}\\), we have\n\\[\na = 1, \\quad b = 2023.\n\\]\n\n\nWe have\n\\[\n2025b^{2026} = 2025 \\cdot 2023^{2026}.\n\\]\nFactor each number:\n\\[\n2025 = 3^4 \\cdot 5^2, \\quad 2023 = 7 \\cdot 17^{2}.\n\\]\nHence,\n\\[\n2025 \\cdot 2023^{2026} = 3^4 \\cdot 5^2 \\cdot 7^{2026} \\cdot 17^{4052}.\n\\]\n\n\n\nIf a number has prime factorization\n\\[\nn = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k},\n\\]\nthen the number of factors is\n\\[\n(\\alpha_1 + 1)(\\alpha_2 + 1) \\cdots (\\alpha_k + 1).\n\\]\n\nHere,\n\\[\nn = 3^4 \\cdot 5^2 \\cdot7^{2026} \\cdot 17^{4052} \\implies \\text{number of factors} = (4+1)(2+1)(2026+1)(4052+1) = 5 \\cdot 3 \\cdot 2027 \\cdot 4053 = \\boxed{123321465}.\n\\]\n\nFinally, the answer is:\n\\[\\boxed{123321465}\n\\]\n\u0026lt;\/p\u0026gt;\n\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9225283}"}]}}}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"2e485d75-e5f8-4e0a-9227-bf58db419e7a","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"7df17827-371e-4aa7-bdf9-db695b4b41e1","items":[{"type":"BlockComponentItem","id":"4380aacf-4cfa-44ea-809b-d1edcd1d138d","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f6a9063d-8170-4066-abf9-ba08cdf16f64","items":[{"type":"HtmlComponent","id":26491659,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page0702026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9222975}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"7 Feb 2026","uid":"7b296c6e-ced8-4315-8bd2-f73db92f1a9a","path":"\/7feb2026","memberOnly":null,"paidMemberOnly":null,"buySpecificProductList":{},"specificTierList":{},"autoPath":false,"authorized":true},{"type":"Page","id":"f_0285f07b-3470-4016-8019-f9845141b9d6","sections":[{"type":"Slide","id":"480f96f8-fbbf-44f5-9db7-dbc1fa0e96ca","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"01d5dcd5-b793-489a-9107-1401c5ff72bb","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"c32c2489-4567-450a-84ce-6ca724b1454b","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"9fcc23c6-d3ef-441b-88a6-1123c7f60b6d","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"e12c0a60-ae1f-41ed-94bc-3c513f55a657","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"c6a1c3f2-593e-4fd3-b13a-523f99a6195c","items":[{"type":"BlockComponentItem","id":"939e01c0-2f0e-47ce-9482-edf224e37623","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"62b41b5b-4934-405f-998c-43887742a6d1","items":[{"type":"HtmlComponent","id":26526656,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem\u0026lt;\/title\u0026gt;\n \u0026lt;script src=\"https:\/\/polyfill.io\/v3\/polyfill.min.js?features=es6\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script id=\"MathJax-script\" async\n src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-mml-chtml.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nJohn selects \\(n\\) distinct integers from the set \\(\\{1, 2, \\dots, 1000\\}\\), where \\(1 \\le n \\le 1000\\), and multiplies them to obtain a product \\(P\\).\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nthe number of ways to choose these integers such that \\(P\\) is divisible by \\(2007\\) can be expressed in the form\n\\[\na^3b - a^2c,\n\\]\nwhere \\(a\\) is as large as possible and \\(a, b, c\\) are positive integers.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nFinally, find the number of positive divisors of\n\\[\na^b \\cdot b^c.\n\\]\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nTo ensure the product \\(P\\) is divisible by \\(2007 = 3^2 \\times 223\\), we divide the set \\(\\{1, 2, \\dots, 1000\\}\\) into these 5 sets:\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;ul\u0026gt;\n \u0026lt;li\u0026gt;\u0026lt;strong\u0026gt;Set 1:\u0026lt;\/strong\u0026gt; Multiples of 3, not 9, excluding 669 (\\(221\\) numbers).\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;\u0026lt;strong\u0026gt;Set 2:\u0026lt;\/strong\u0026gt; Multiples of 9 (\\(111\\) numbers).\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;\u0026lt;strong\u0026gt;Set 3:\u0026lt;\/strong\u0026gt; \\(\\{223, 446, 892\\}\\) (\\(3\\) numbers).\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;\u0026lt;strong\u0026gt;Set 4:\u0026lt;\/strong\u0026gt; \\(\\{669\\}\\) (\\(1\\) number).\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;\u0026lt;strong\u0026gt;Set 5:\u0026lt;\/strong\u0026gt; All other numbers (\\(664\\) numbers).\u0026lt;\/li\u0026gt;\n\u0026lt;\/ul\u0026gt;\n\n\u0026lt;h3\u0026gt;Casework\u0026lt;\/h3\u0026gt;\n\n\u0026lt;p\u0026gt;\n\u0026lt;strong\u0026gt;Case 1: Including 669.\u0026lt;\/strong\u0026gt; Since \\(669 = 3 \\times 223\\), we only need at least one more factor of 3 from the remaining \\(999\\) integers.\n\\[ \\text{Ways}_1 = 2^{999} - 2^{667} \\]\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\n\u0026lt;strong\u0026gt;Case 2: Excluding 669.\u0026lt;\/strong\u0026gt; We must choose at least one from Set 3 (\\(2^3 - 1 = 7\\) ways).\n\u0026lt;\/p\u0026gt;\n\u0026lt;ul\u0026gt;\n \u0026lt;li\u0026gt;Case 2.1: Including at least one number from Set 2: \\(7 \\times (2^{111} - 1) \\times 2^{885}\\)\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;Case 2.2: Excluding Set 2, but including at least two from Set 1: \\(7 \\times (2^{221} - 222) \\times 2^{664}\\)\u0026lt;\/li\u0026gt;\n\u0026lt;\/ul\u0026gt;\n\n\u0026lt;h3\u0026gt;Final Result\u0026lt;\/h3\u0026gt;\n\n\u0026lt;p\u0026gt;\nSumming all cases:\n\\[ (2^{999} - 2^{667}) + (7 \\cdot 2^{996} - 7 \\cdot 2^{885}) + (7 \\cdot 2^{885} - 1554 \\cdot 2^{664}) \\]\n\\[ = 15 \\cdot 2^{996} - 1562 \\cdot 2^{664} \\]\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nComparing this to the form \\(a^3b - a^2c\\) where \\(a\\) is as large as possible:\n\\[ a = 2^{332}, \\quad b = 15, \\quad c = 1562 \\]\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nThe number of positive divisors of \\(a^b \\cdot b^c = (2^{332})^{15} \\cdot (3 \\cdot 5)^{1562} = 2^{4980} \\cdot 3^{1562} \\cdot 5^{1562}\\) is:\n\\[ (4980+1)(1562+1)(1562+1) \\]\n\\[ 4981 \\cdot 1563 \\cdot 1563 = \\boxed{12168428589} \\]\n\u0026lt;\/p\u0026gt;\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9225281}"}]}}}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"59afc29a-dc73-4fb4-a7b8-f86a4929d3a7","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"e7951618-516e-4187-be55-8b6dce681d88","items":[{"type":"BlockComponentItem","id":"8fe44345-4644-4b38-b1dd-ac4fead9460a","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"bc92c894-2c4c-4e3f-b726-8115ef44594b","items":[{"type":"HtmlComponent","id":26526639,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page14022026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9225279}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"14 Feb 2026","uid":"7969d92a-65b9-42f6-a5fe-6ff20c269c48","path":"\/14feb2026","memberOnly":null,"paidMemberOnly":null,"buySpecificProductList":{},"specificTierList":{},"autoPath":false,"authorized":true},{"type":"Page","id":"f_bee8f409-b952-4ac5-b127-f813e4d39d8b","sections":[{"type":"Slide","id":"f_9346c81f-0334-4a94-8b83-7e02701ac8db","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_40571de8-b012-4211-9771-b2a6039fe509","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_73e2c255-ad24-4a1d-b4cc-a1ac7e976e38","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_7e8c625a-a448-48ec-b545-0efe4b955be2","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_d32568ce-198b-42c5-ac39-d3d5377e611e","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_07172bec-e08a-4a3f-a571-5c70fcf8ba30","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_cfa0d21a-58a9-4dfb-a499-9202dedc19b2","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_f9eb1bf1-b600-4fa3-9fd8-035ff655c4fb","defaultValue":null,"items":[{"type":"HtmlComponent","id":26575746,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html\u0026gt;\n\u0026lt;head\u0026gt;\n\u0026lt;meta charset=\"utf-8\"\u0026gt;\n\u0026lt;title\u0026gt;Math Problem\u0026lt;\/title\u0026gt;\n\n\u0026lt;script\u0026gt;\nwindow.MathJax = {\n tex: {inlineMath: [['

  • —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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, '

  • —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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], ['\\\\(', '\\\\)']]},\n svg: {fontCache: 'global'}\n};\n\u0026lt;\/script\u0026gt;\n\n\u0026lt;script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nIn the Cartesian coordinate plane $xOy$, consider the parabola\n$\nC: y = x^2\n$\nand the line\n$\n\\ell: y = 13x + 44.\n$\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nThe line $\\ell$ intersects the parabola $C$ at two points $P$ and $Q$, \nwith $P$ lying to the left of $Q$.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nLet $A, B, C$ be arbitrary points on the arc of the parabola between $P$ and $Q$.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nRotate the region of the second and third quadrants about the $y$-axis \nby $120^\\circ$ into three-dimensional space.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nLet $O$ denote the origin. \n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nIf the maximum volume of $OABC$ can be expressed in the form\n$\n\\frac{a\\sqrt{b} + c\\sqrt{d}}{e},\n$\nwhere $a,b,c,d,e$ are positive integers, \n$\\gcd(a,c,e)=1$, and $b,d$ are square-free integers, \nfind\n$\na+b+c+d+67e\n$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;hr\u0026gt;\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n\n\n\n\u0026lt;p\u0026gt;The intersection of the parabola $y = x^2$ and the line $y = 13x + 44$ is determined by solving:\u0026lt;\/p\u0026gt;\n$x^2 - 13x - 44 = 0$\n\u0026lt;p\u0026gt;Using the quadratic formula:\u0026lt;\/p\u0026gt;\n$x = \\frac{13 \\pm \\sqrt{169 + 176}}{2} = \\frac{13 \\pm \\sqrt{345}}{2}$\n\u0026lt;p\u0026gt;Let $x_1 = \\frac{13 - \\sqrt{345}}{2}$ and $x_2 = \\frac{13 + \\sqrt{345}}{2}$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;h3\u0026gt;Casework\u0026lt;\/h3\u0026gt;\n\u0026lt;p\u0026gt;Points $A, B,$ and $C$ lie on the arc of the parabola. We analyze the volume based on their position relative to the origin $O$ after a $120^\\circ$ rotation about the $y$-axis.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 1:\u0026lt;\/strong\u0026gt; If $A, B, C$ are all to the left of the $y$-axis ($x \u0026lt; 0$), the points remain coplanar with the origin. The volume of $OABC$ is $0$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 2:\u0026lt;\/strong\u0026gt; If $A, B, C$ are all to the right of the $y$-axis ($x \u0026gt; 0$), the points remain coplanar with the origin. The volume of $OABC$ is $0$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 3:\u0026lt;\/strong\u0026gt; Point $A$ is to the left of $O$ \n\u0026lt;p\u0026gt;In this configuration, we place point $A$ at the left intersection $x_1$ and points $B$ and $C$ on the right side of the arc. To find the area of the base $\\triangle OBC$, we define a point $Q$ directly below $B$ on the line $\\ell$ connecting $O$ and $C$.\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;1. Area Decomposition:\u0026lt;\/strong\u0026gt;The area of $\\triangle OBC$ can be split into two triangles, $\\triangle OBQ$ and $\\triangle CBQ$, by the vertical segment $BQ$.\u0026lt;\/p\u0026gt;\u0026lt;ul\u0026gt;\u0026lt;li\u0026gt;For $\\triangle OBQ$, the base is $BQ$ and the height is the horizontal distance $x_B$. Thus, $\\text{Area}_{OBQ} = \\frac{x_B \\cdot BQ}{2}$.\u0026lt;\/li\u0026gt;\u0026lt;li\u0026gt;For $\\triangle CBQ$, the base is $BQ$ and the height is the horizontal distance $(x_C - x_B)$. Thus, $\\text{Area}_{CBQ} = \\frac{(x_2 - x_B) \\cdot BQ}{2}$.\u0026lt;\/li\u0026gt;\u0026lt;\/ul\u0026gt;\u0026lt;p\u0026gt;Summing these areas, we get the total base area:$\\text{Area}_{OBC} = \\frac{x_B \\cdot BQ}{2} + \\frac{(x_2 - x_B) \\cdot BQ}{2} = \\frac{x_2 \\cdot BQ}{2}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;2. Maximizing $BQ$:\u0026lt;\/strong\u0026gt;The length $BQ$ is the vertical distance between the parabola $y=x^2$ and the segment $OC$. The line $OC$ has the equation $y = \\frac{x_2^2}{x_2}x = x_2 x$. Therefore:$BQ = x_2 x_B - x_B^2$This quadratic is maximized when $x_B = \\frac{x_2}{2}$. At this point, $BQ = x_2(\\frac{x_2}{2}) - (\\frac{x_2}{2})^2 = \\frac{x_2^2}{4}$.Substituting this back into the area formula:$\\text{Area}_{OBC} = \\frac{x_2}{2} \\cdot \\frac{x_2^2}{4} = \\frac{x_2^3}{8}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;3. Volume and Vieta's Theorem:\u0026lt;\/strong\u0026gt;The height of the tetrahedron in the rotated 3D space is $h = -x_1 \\frac{\\sqrt{3}}{2}$ (accounting for the $120^\\circ$ rotation). The volume is:$V = \\frac{1}{3} \\cdot \\left( \\frac{x_2^3}{8} \\right) \\cdot \\left( \\frac{-x_1\\sqrt{3}}{2} \\right) = \\frac{-x_1 x_2^3 \\sqrt{3}}{48}$By Vieta's Theorem for the equation $x^2 - 13x - 44 = 0$, we know $x_1 x_2 = -44$. We can rewrite the volume as:$V = \\frac{-(x_1 x_2) x_2^2 \\sqrt{3}}{48} = \\frac{-(-44) x_2^2 \\sqrt{3}}{48} = \\frac{44 \\sqrt{3} x_2^2}{48} = \\frac{11 \\sqrt{3} x_2^2}{12}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;Using $x_2^2 = \\frac{257 + 13\\sqrt{345}}{2}$:$V = \\frac{11\\sqrt{3}}{12} \\cdot \\frac{257 + 13\\sqrt{345}}{2} = \\frac{2827\\sqrt{3} + 143\\sqrt{1035}}{24} = \\frac{2827\\sqrt{3} + 429\\sqrt{115}}{24}$\u0026lt;\/p\u0026gt;\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 4:\u0026lt;\/strong\u0026gt; Point $A$ is to the right of $O$, while points $B$ and $C$ are to the left. The volume is $\\frac{11\\sqrt{3}x_1^2}{12}$. The resulting volume is:\n$V = \\frac{2827\\sqrt{3} - 429\\sqrt{115}}{24}$\n\u0026lt;\/p\u0026gt;\n\n\n\u0026lt;p\u0026gt;Comparing Case 3 and Case 4, the maximum volume is given by Case 3. Expressing this in the form $\\frac{a\\sqrt{b} + c\\sqrt{d}}{e}$:\u0026lt;\/p\u0026gt;\n\u0026lt;ul\u0026gt;\n \u0026lt;li\u0026gt;$a = 2827, b = 3$\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;$c = 429, d = 115$\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;$e = 24$\u0026lt;\/li\u0026gt;\n\u0026lt;\/ul\u0026gt;\n\n\u0026lt;p\u0026gt;The final value is:\n$a + b + c + d + 67e = 2827 + 3 + 429 + 115 + 1608 = \\boxed{4982}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9227546}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_818046a6-1d5a-482a-a00f-af2972f97232","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_41ef131a-26e2-46ac-a695-9ffdc390633a","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"1b977d06-058c-489c-8c46-80d794f6ad0d","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"80a831dc-5c11-4696-a6fa-789cb8130130","items":[{"type":"BlockComponentItem","id":"7c7a9a30-a9e2-4974-b807-183fc8044f48","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"b20b5f62-f8a2-4cc9-be3e-476e2c369434","items":[{"type":"HtmlComponent","id":26575749,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page21022026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9227547}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"21 Feb 2026","uid":"58c69a8a-09d7-4b77-a88f-ba5fa2ae0ff9","path":"\/21feb2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_26fa4fa7-bd57-46bd-b54b-3959bb0e5ea2","sections":[{"type":"Slide","id":"f_0cc98249-fbce-4d51-86f2-966397a3e4b6","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_b390b2e2-9fc8-417d-9403-798ccd270c2f","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_8e4d2a35-ec8c-4b81-b7e9-887e7bc76565","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_9ee279b7-4516-45b3-bb79-d0700c871ce2","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_bb12efdb-8eb7-4dea-aa2c-59471cff8cf2","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_25dcbfd2-84b0-4796-8608-32ca4102ae40","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_3c21012a-7fe5-4d38-a21a-9c5b416f36bc","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_5cc860cc-b599-4700-a501-4275e781569b","defaultValue":null,"items":[{"type":"HtmlComponent","id":26629714,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Geometry Problem\u0026lt;\/title\u0026gt;\n\n \u0026lt;!-- MathJax --\u0026gt;\n \u0026lt;script\u0026gt;\n window.MathJax = {\n tex: {\n inlineMath: [['

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    Let's Solve Problems!

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], ['\\\\(', '\\\\)']],\n displayMath: [['$', '$'], ['\\\\[', '\\\\]']]\n }\n };\n \u0026lt;\/script\u0026gt;\n \u0026lt;script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-chtml.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\n\u0026lt;body\u0026gt;\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nFind the minimum $p \\in \\mathbb{Z}^+$ such that:$20151121^{20151121} \\mid p!$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;1. Base Factorization:\u0026lt;\/strong\u0026gt;\n The number $20151121$ is a perfect power:\n $20151121 = 4489^2 = (67^2)^2 = 67^4$\n Substituting this back into the expression:\n $(67^4)^{20151121} = 67^{80604484}$\n We must find the smallest $p$ such that $p!$ contains at least $80604484$ factors of $67$.\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;2. Legendre's Formula:\u0026lt;\/strong\u0026gt;\n The exponent of a prime $q$ in $p!$ is given by:\n $E_q(p!) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\frac{p}{q^k} \\right\\rfloor$\n Using the approximation $E_q(p!) \\le \\frac{p}{q-1}$:\n $\\frac{p}{66} \\ge 80604484 \\implies p \\ge 5319895944$\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;3. Conclusion:\u0026lt;\/strong\u0026gt;\n Testing the multiple $p = 5319895944$ yields an exponent of $80604483$. Adding the next multiple of $67$ gives:\n $p = \\boxed{5319896011}$\n \u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9233160}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_1ea681e2-9a55-4853-af88-e410e9da2ab0","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_213f30d9-73b7-4db4-b744-31a283a563d3","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"c1790b81-d45d-40ad-a72e-5ab1ee5d1c54","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"057ed604-fb62-4b46-87c8-0c8f9dbf36d2","items":[{"type":"BlockComponentItem","id":"0743316f-81d6-4b65-bbc1-01e25bb95d3a","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"914c5561-29b6-41bd-988f-1bef731b1429","items":[{"type":"HtmlComponent","id":26629720,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page28022026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9233161}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"28 Feb 2026","uid":"6eac4717-e233-4df6-9016-2d3570e0cb51","path":"\/28feb2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_222a9ce6-a4b8-4160-b3a7-597b9854b917","sections":[{"type":"Slide","id":"f_b21f4940-d576-42f1-a034-274a8eaa3c3a","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_bcf267cb-93f9-45de-ad3e-29a6d8323673","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_bb747b57-cb34-480f-a8bd-45d038862fcb","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_0936d558-a236-4b33-86db-6e7a7e6f067a","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_f8a601e4-0555-4ab6-a72f-7d79fa4964d5","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_1f9c2f49-2991-461d-a906-13122404f003","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_406bb2dc-d82b-4bef-a299-00883323bdb6","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_de862e98-5119-480f-a31c-624af29d5978","defaultValue":null,"items":[{"type":"HtmlComponent","id":26676340,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n\u0026lt;meta charset=\"UTF-8\"\u0026gt;\n\u0026lt;title\u0026gt;Math Problem: Tangents through a Point\u0026lt;\/title\u0026gt;\n\u0026lt;!-- KaTeX CSS --\u0026gt;\n\u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\n onload=\"renderMathInElement(document.body);\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;style\u0026gt;\n body { font-family: Arial, sans-serif; padding: 20px; line-height: 1.6; }\n h1 { color: #2c3e50; }\n .formula { font-size: 1.2em; margin: 10px 0; }\n\u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;Consider the cubic function:\u0026lt;\/p\u0026gt;\n\n \\( y = x^3 - 11x^2 + 26x - 51 \\)\n\n\n\u0026lt;p\u0026gt;Let \\( Q(4, y_0) \\) be a point. There exist exactly three tangents to the curve that pass through \\( Q \\).\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;The range of \\( y_0 \\) is \\((a,b)\\). \\( b-a \\) can be written as a reduced fraction \\(\\frac{c}{d}\\) with coprime integers \\(c\\) and \\(d\\), and then compute:\u0026lt;\/p\u0026gt;\n\n \\( 80c - 3d \\)\n\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;Let $(t, f(t))$ be the point of tangency on the curve $y = x^3 - 11x^2 + 26x - 51$The derivative is:\u0026lt;\/p\u0026gt;\n $ f'(t) = 3t^2 - 22t + 26 $\n \u0026lt;p\u0026gt;The equation of the tangent line passing through $Q(4, y_0)$ and $(t, f(t))$ is given by the slope formula:\u0026lt;\/p\u0026gt;\n $ \\frac{f(t) - y_0}{t - 4} = f'(t) $\n \u0026lt;p\u0026gt;Substituting the functions and simplifying:\u0026lt;\/p\u0026gt;\n $ (t^3 - 11t^2 + 26t - 51) - y_0 = (3t^2 - 22t + 26)(t - 4) $\n $ t^3 - 11t^2 + 26t - 51 - y_0 = 3t^3 - 34t^2 + 114t - 104 $\n $ 2t^3 - 23t^2 + 88t - 53 - y_0 = 0 $\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;For exactly three tangents, the cubic $g(t) = 2t^3 - 23t^2 + 88t - 53 - y_0$ must have three distinct real roots. This occurs when the local extrema of g(t) lie on opposite sides of the t-axis.\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;Find critical points via $g'(t) = 0$\u0026lt;\/p\u0026gt;\n $ 6t^2 - 46t + 88 = 0 \\implies 3t^2 - 23t + 44 = 0 $\n $ (3t - 11)(t - 4) = 0 $\n \u0026lt;p\u0026gt;The critical points are at $t = \\frac{11}{3}, 4$\u0026lt;\/p\u0026gt;\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;We require $g(4) \\cdot g\\left(\\frac{11}{3}\\right) \u0026lt; 0$ Evaluating the function at the critical points:\u0026lt;\/p\u0026gt;\n \u0026lt;ul\u0026gt;\n $g(4) = 59 - y_0$\n $g\\left(\\frac{11}{3}\\right) = \\frac{1594}{27} - y_0$\n \u0026lt;\/ul\u0026gt;\n \u0026lt;p\u0026gt;This gives the range $ y_0 \\in \\left( 59, \\frac{1594}{27} \\right) $ so $a = 59$ $b = \\frac{1594}{27}$\u0026lt;\/p\u0026gt;\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;The difference is:\u0026lt;\/p\u0026gt;\n $ b - a = \\frac{1594}{27} - \\frac{1593}{27} = \\frac{1}{27} $\n \u0026lt;p\u0026gt;Where $c = 1$ $d = 27$ The final value is:\u0026lt;\/p\u0026gt;\n \u0026lt;div class=\"result-box\"\u0026gt;\n $ 80c - 3d = 80(1) - 3(27) = 80 - 81 = \\boxed{-1} $\n \u0026lt;\/div\u0026gt;\n \u0026lt;\/div\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9238442}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_cf362c1b-7e51-4860-8b64-e3b988ba5bf7","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_0798bcd4-9136-4f21-b05c-9ad25d1dd370","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"abc35b75-0cc0-4d7f-8936-391355245793","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"5e3e5b10-a19e-4f67-aa26-65e1ffb7d6f5","items":[{"type":"BlockComponentItem","id":"2986dc0b-a28f-458e-9474-e1e67b00e232","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"1bca910c-18ec-4d22-92c7-0960cfb61ddd","items":[{"type":"HtmlComponent","id":26676348,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page07032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9238443}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"7 Mar 2026","uid":"69777895-0258-4183-a297-6c832978fb48","path":"\/7mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_61742f60-2f5c-46a5-a074-5ba1309341b5","sections":[{"type":"Slide","id":"f_4cc583f4-8de3-4d49-a20f-541eec20631c","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_943c3c83-91c1-4c45-9f34-9883ff01ebe8","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_cb85c8c0-e3be-4d13-9ce6-fb0bbd975f4c","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_fcc879bc-7a3e-41f8-adeb-716bcaec40bf","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_aca23767-705b-40da-9dd4-a63576daeea3","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_b5ca18a4-9070-4063-bf1e-3a5ecc636fa1","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_6419c6c2-7f9d-4848-b2b1-2780c6b3adb8","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_00454226-f64b-4cdc-b208-20cae5e8fb56","defaultValue":null,"items":[{"type":"HtmlComponent","id":26721757,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n\u0026lt;meta charset=\"UTF-8\"\u0026gt;\n\u0026lt;title\u0026gt;Math Problem: Tangents through a Point\u0026lt;\/title\u0026gt;\n\u0026lt;!-- KaTeX CSS --\u0026gt;\n\u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\n onload=\"renderMathInElement(document.body);\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;style\u0026gt;\n body { font-family: Arial, sans-serif; padding: 20px; line-height: 1.6; }\n h1 { color: #2c3e50; }\n .formula { font-size: 1.2em; margin: 10px 0; }\n\u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\n\u0026lt;div class=\"problem-box\"\u0026gt;\n \u0026lt;p\u0026gt;\n Kevin selects three distinct integers $a, b, c$ from the set:\n \u0026lt;\/p\u0026gt;\n $ \\{1, 2, 3, \\dots, 2026\\} $\n \u0026lt;p\u0026gt;\n The probability that these numbers can form the side lengths of a triangle with non-zero area can be written as $\\frac{a}{b}$ which a and b are coprime. Find a+b.\n \u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\u0026lt;div class=\"solution-box\"\u0026gt;\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 1: Define the Sample Space.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n We are choosing 3 distinct integers from the set $\\{1, 2, \\dots, 2026\\}$The total number of ways is:\u0026lt;\/p\u0026gt;\n $ S = \\binom{2026}{3} = \\frac{2026 \\times 2025 \\times 2024}{6} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 2: Apply the Triangle Inequality.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n For three numbers $x \u0026lt; y \u0026lt; z$ to form a triangle, we must have $x + y \u0026gt; z$. The number of such combinations $T_n$ for an even $n$ is given by:\u0026lt;\/p\u0026gt;\n $ T_n = \\frac{n(n-2)(2n-5)}{24} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 3: Calculate the Probability.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n simplifies algebraically to:\u0026lt;\/p\u0026gt;\n $ P = \\frac{T_n}{S} = \\frac{2n - 5}{4(n - 1)} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 4: Substitute $n = 2026$.\u0026lt;\/span\u0026gt;\u0026lt;\/p\u0026gt;\n $ P = \\frac{2(2026) - 5}{4(2026 - 1)} = \\frac{4052 - 5}{4(2025)} = \\frac{4047}{8100} = \\frac{1349}{2700} $\n\n \n \n $ a + b = 1349 + 2700 = \\boxed{4049} $\n \u0026lt;\/div\u0026gt;\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9242502}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_e05d8fde-c33f-4aa7-836c-c86fb143074c","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_7ca839d8-a57d-4485-ab82-e2373f37215b","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"9c41d104-c15a-4fea-96ce-c3a21cf39cf5","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"2479ead5-73e4-48f8-8a2b-5407e9128f8f","items":[{"type":"BlockComponentItem","id":"8232dee5-dcea-4031-ac82-1a6faefd0674","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"9e6c5106-06d6-418f-847a-358d42062ddb","items":[{"type":"HtmlComponent","id":26721758,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page14032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9242503}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"14 Mar 2026","uid":"13112462-6615-4c8a-bb4d-23b3d269185f","path":"\/14mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_f86169c2-6be2-4cfc-90b3-6b2fe771fffa","sections":[{"type":"Slide","id":"f_7ad77ed3-5348-49ed-861d-9589ccbce485","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_cd49bb59-b76a-46b8-ad00-3069f1d0c008","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_dd55aaec-8b57-428b-97e1-38da8e90ac96","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_a2ffb588-57af-4302-8162-9ed41f1d5e14","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_eeaa4ca6-6c71-47b2-8144-30c27a291fad","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_0729e330-afad-4977-87df-9c3029ead4b4","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_14efc80a-506a-48cd-9670-44d100dbe8b7","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_5ddb0c18-d876-403c-b8fa-96fcc63a0915","defaultValue":null,"items":[{"type":"HtmlComponent","id":26815313,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

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    Let's Solve Problems!

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  • This Week's Problem

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \u0026lt;p\u0026gt;\n A $Q$-digit number in base $P$ is said to possess Property $(P, Q)$ if all its digits are distinct, and the absolute difference between any two of its digits does not equal any digit present in the number itself.\n \u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;\n Let $S(P, Q)$ be the set of all numbers possessing Property $(P, Q)$. There exists a positive integer $T$ such that $S(T, 2026)$ is non-empty, while for all positive integers $R \u0026lt; T$, the set $S(R, 2026)$ is empty.\n \u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;\n If $M$ is the smallest element in $S(T+2028, 2027)$, find the value of:\n \u0026lt;\/p\u0026gt;\n \n \n $ (M + T) \\pmod{2026} $\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \u0026lt;p\u0026gt;\n 1. Prove that $T \u0026gt; 4051$:\n We set the digits of the $2026$-digits number is $a_1, a_2, a_3 ... a_{2026}(a_1 \u0026lt; a_2 \u0026lt; a_3 ... \u0026lt; a_{2026})$, and the biggest digit is $a_{2026}$.\n\nConstruct a sequence $b_1, b_2, b_3 ... b_{2025}(b_k = a_{2026} - a_k)$ If there is two distinct positive integers $i,j$, such that $a_i = b_j$. We can find that $a_i + a_j = a_{2026}$\n\n \nThus, we have to make that $a_1, a_2, a_3 ... a_{2026}, b_1, b_2 ... b_{2025}$ $4051 $ elements are all distinct. When $T \\le 4051$, there are at least two same numbers.\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n 2. Construct an example when $T = 4052$:\n $\na_1 = 1,\na_2 = 3,\na_3 = 5,\na_4 = 7\n...\na_{2026} = 4051\n$\nIt is easy to prove that it has Property$(4052,2026)$.\n\u0026lt;\/p\u0026gt;\n 3. Find the smallest number that has Property$(6080,2027)$ and final answer:\nSimilarly,$a_1 = 1, a_2 = 3 ... a_{2027} = 4053$.\nWe just need to find that \n$\\sum_{i=1}^{2027} 6080^{2027-i}a_i \\equiv \\sum_{i=1}^{2027} 2^{2027-i}(2i-1) \\equiv 3 \\cdot 2^{2027} - 4057 \\pmod{2026}$\n\n $(M+T) \\equiv (3 \\cdot 2^{2027} - 5) \\pmod{2026}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nBy Fermat's Little Theorem, \n We can find that \n$2^{2027} \\equiv 8 \\pmod{1013}$ and $2^{2027} \\equiv 8$ or $ 1021 \\pmod{2026}$\n\nBecause $2^{2027}$ is even, $2^{2027} \\equiv 8 \\pmod{2026}$\n\nThus, $(M+T) \\bmod{2026} = 3 \\cdot 8 - 5 = \\boxed{19}$\n\n\n \u0026lt;\/p\u0026gt;\n\n \n \n \n \n \n\u0026lt;\/div\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9256577}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_2bc385b4-5720-4ea8-b0d2-ccc8d539c584","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_c06c283c-c5e4-4541-b9d1-21c05ea462e7","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"27a85022-f4cc-44e7-9abe-3f7269863aae","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"5bb2e3ea-7b09-4799-96be-e30cc77b9335","items":[{"type":"BlockComponentItem","id":"f7b01d89-2444-442f-9fd5-b8a67a337e82","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"dc8cfb7d-22d5-44c7-bc3c-49167ee60a55","items":[{"type":"HtmlComponent","id":26815445,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page21032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9256585}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"21 Mar 2026","uid":"2cefb69d-ee7f-415c-854d-4ae03989254d","path":"\/21mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_5e7b01b0-8f45-43dd-8e25-a2845eed1454","sections":[{"type":"Slide","id":"f_43e4e8bc-b8e6-4e2a-b460-a66939bfccbb","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_78dc2f4b-0657-4d8d-a849-5fd6d644141a","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_5fa03bc4-5d00-42ea-9b4d-6126eddc806e","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_ec3a2983-9cfd-4406-8b82-a32d09d8a0d9","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_8d073596-d408-4720-9407-5cd78b180180","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_efaef965-bade-41d4-8d5d-98f7fcd87cdf","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_0875ebff-459c-4cf1-8d1b-b3bd8fc243a0","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_72fc710d-7b0e-4a42-8fc1-5128de336ce4","defaultValue":null,"items":[{"type":"HtmlComponent","id":26860037,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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    Let's Solve Problems!

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  • This Week's Problem

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;\n Let $A$ be a set of all right circular cones such that each cone in $A$ has an inscribed sphere with a radius of $1000$.From the set $A$, let $P$ be the cone that has the minimum volume. An inscribed sphere of radius $1000$ is placed inside cone $P$.Let $R$ be a positive integer, and let $M$ be a positive integer such that $M$ spheres, each of radius $R$, can be placed simultaneously into the gap at the bottom of the cone (the region bounded by the cone's base, its lateral surface, and the large inscribed sphere).\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;Find the value of $RM+R+M$ when the integer $R$ is at its maximum possible value.\u0026lt;\/p\u0026gt;\n \u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;p\u0026gt;\n 1. find the smallest value of the volume:\n\u0026lt;p\u0026gt;$V = \\frac{\\pi r^2 h}{3}$ , which $s$ represents the base radius of the cone\u0026lt;\/p\u0026gt;\nWe can take a vertical section from the cone.\nThe section is an isosceles triangle, with the $2r$ base side and the $h$ height. \nIt is easy to prove that $2arctan \\frac{1000}{r} = arctan \\frac{h}{r}$\n\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nThus, $h = \\frac{2000r^2}{r^2-1000000}$, and plug it in to $V$ to get $V = \\frac{2000\\pi r^4}{3r^2-3000000}$ and $V' = \\frac{4000\\pi r^3(r^2 - 2000000)}{3(r^2 - 1000000)}$, when $r = 1000\\sqrt{2}$, the value of the volume is smallest.\n\n\n \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n2. find the biggest radius of the balls:\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nConsider a right triangle $ABC$, with $C = \\frac{\\pi}{2}, AC = 1000\\sqrt{2}, BC = 4000$, and $D$ is on the side $BC$, with $CD = 1000$. Connect $AD$. Let $x$ and point$P, M$ on $AD$ be the radius ,center and a point on segment $AD$ of the small ball, respectively, which $PA = \\sqrt{3}x, PM = x, AM = (1+\\sqrt{3})x \\le 1000(\\sqrt{3}-1)$\nThus, $x \\le 1000(2-\\sqrt{3}) \\approx 267.9492$, which means $x = 267 = R$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n3. find the biggest number of the balls:\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nWhen $R = 267$, let the distance between the center of the ball and the central axis of the cone be $d$. where $d \\le (1000-267)\\sqrt{2} \\approx 1036.62$.\nThus the number of balls is $M = \\lfloor \\frac{\\pi}{arcsin \\frac{R}{d}} \\rfloor = 12$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nFinally, the answer is $RM + R + M = 267 \\cdot 12 + 267 + 12 = \\boxed{3483}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9258844}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_2134036c-79b1-461c-b055-f69f32352033","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_bc2b9845-108d-44f6-b386-c46792d6b415","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"9c3f5b42-2788-42bb-bed8-9bfed3f6fab4","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"65f65d2b-868c-46c9-b749-a1544af43526","items":[{"type":"BlockComponentItem","id":"5fa6db7e-827a-4bc4-b789-491b3dac330f","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"2ec0fc2c-a7f9-4e9c-af35-f156fc637f68","items":[{"type":"HtmlComponent","id":26860185,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page28032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9258847}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"28 Mar 2026","uid":"01f4b56f-deea-4032-8e6b-9a8e829de6a4","path":"\/28mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_031324ca-5db8-4209-9e29-58f1d93c68e7","sections":[{"type":"Slide","id":"f_952da0aa-1376-4bbc-b3bd-662077b509dd","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_60e46bec-c7c3-4b51-b314-a0e8a5796cc4","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_8626653f-47f9-44e8-8e9c-4320ac5e0f67","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_b3a6abc0-5da9-4974-84fa-60b1d813ce87","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_8154881e-dae4-4962-8e95-2cbf15c89248","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_dec7988f-35f0-4e5d-994f-800e500611e4","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_bbd1a58d-8820-4304-a803-17e9f29e4974","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_4b5c1bb1-dfe7-4adb-a9e6-a9e8849092e4","defaultValue":null,"items":[{"type":"HtmlComponent","id":26905642,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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  • This Week's Problem

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    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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  • This Week's Problem

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    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;\n Let $S = \\{a_1, a_2, a_3, \\dots, a_{2026}\\}$ be a set of $2026$ real numbers.For each $k \\in \\{1, 2, \\dots, 2026\\}$, let $S_k$ be the subset of $2025$ elements obtained by removing $a_k$ from $S$. It is given that the standard deviation of the elements in $S_k$ is equal to $k$. If $D$ is the variance (the square of the standard deviation) of the original set $S$ of all $2026$ numbers, and $D$ can be written as $\\frac{a}{b}$, where both a and b are positive integers and coprime. Find $a+b$.\u0026lt;\/p\u0026gt;\n \u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n \u0026lt;p\u0026gt;\n1. Establish the relationship between the original variance and the subset variances:\n\u0026lt;\/p\u0026gt;\n\nLet $n = 2026$, $\\mu$ be the mean of set $S$, and $D$ be its variance. \nThe definition of variance gives $\\sum_{i=1}^{n} (a_i - \\mu)^2 = nD$.\nFor each $k$, the mean of the subset $S_k$ is $\\mu_k = \\frac{n\\mu - a_k}{n-1}$.\nThe variance of $S_k$ is given as $\\sigma_k^2 = k^2$. \nUsing the computational formula for variance:\n$(n-1)k^2 = \\sum_{i \\neq k} a_i^2 - (n-1)\\mu_k^2$\n\nSubstituting $\\sum a_i^2 = nD + n\\mu^2$ and the expression for $\\mu_k$, we can derive the identity:\n$(n-1)^2 k^2 = n(n-1)D - n(a_k - \\mu)^2$\n\n\u0026lt;p\u0026gt;\n2. Sum over all $k$ to eliminate individual elements:\n\u0026lt;\/p\u0026gt;\n\nSumming the above expression from $k=1$ to $n$:\n$(n-1)^2 \\sum_{k=1}^{n} k^2 = \\sum_{k=1}^{n} \\left[ n(n-1)D - n(a_k - \\mu)^2 \\right]$\nApplying the sum of squares formula $\\sum_{k=1}^{n} k^2 = \\frac{n(n+1)(2n+1)}{6}$ and the fact that $\\sum (a_k - \\mu)^2 = nD$:\n$(n-1)^2 \\frac{n(n+1)(2n+1)}{6} = n^2(n-1)D - n(nD)$\nSimplifying the right side: $n^2(n-1)D - n^2D = n^2(n-2)D$.\n\n\u0026lt;p\u0026gt;\n3. Find the value of $D$ and $a+b$:\n\u0026lt;\/p\u0026gt;\n\nThus, $D = \\frac{(n-1)^2(n+1)(2n+1)}{6n(n-2)}$. Plug in $n = 2026$:\n$D = \\frac{2025^2 \\cdot 2027 \\cdot 4053}{6 \\cdot 2026 \\cdot 2024}$\nSince $4053 = 3 \\times 1351$, we simplify the fraction:\n$D = \\frac{2025^2 \\cdot 2027 \\cdot 1351}{2 \\cdot 2026 \\cdot 2024}$\nLet this simplified fraction be $\\frac{11229467248125}{8201248}$. \nPerforming the final arithmetic gives the coprime integers $a$ and $b$.\n\nFinally, the answer is $a+b = \\boxed{11229475449373}$\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9265262}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_71ff5d25-39d4-4f4c-8370-b1dd99146982","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_03f6bc38-f3cb-41c6-b501-4851a0709d6b","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"f6e71291-cceb-40ae-bcde-720792c84fac","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"98f6b60a-97cb-4684-90c4-c9a405529eae","items":[{"type":"BlockComponentItem","id":"9f427068-e060-4c24-a3bd-3939c2867801","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"aadbbb16-be32-4fca-b17e-a5213e95bd0e","items":[{"type":"HtmlComponent","id":26989941,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page04042026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270810}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"4 Apr 2026","uid":"f1da7ee1-9a4c-400f-86ad-1d337bb4d180","path":"\/4apr2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_bbcfb68e-e238-4f35-a0dc-7bcf7da0d67d","sections":[{"type":"Slide","id":"f_68756df4-a642-4f85-be53-9bc04418a190","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_d7378037-93da-4e3c-a5db-72068e4f80b2","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_e57fc25c-8005-4db5-a201-763d6b6b01d0","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_837767d2-5c2d-40a3-84b1-2481d1aa44bf","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_13f3bec7-854d-4bc6-9b9a-7d5118d47a7c","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_9e231b4e-72fc-47cd-8128-96e82de94fb2","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_ce5238fc-1a96-4c51-a21a-cb60614a77a0","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_0ab2ea1e-bc66-4cf3-bc28-eeec6f830e23","defaultValue":null,"items":[{"type":"HtmlComponent","id":26947587,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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    Let's Solve Problems!

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;Hogwarts School of Witchcraft and Wizardry has two types of spells. Spell $1$ increases the number on the blackboard by $A$, and Spell $2$ increases it by $B$. If the initial number on the blackboard is $1$, and both $A$ and $B$ are positive integers, how many pairs $(A, B)$ exist such that the number on the blackboard reaches exactly $6517$ after exactly $6$ spells (using each spell at least once)?\u0026lt;\/p\u0026gt;\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;p\u0026gt;There are FIVE different cases:\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 1: We used Spell $1$ 1 time and Spell $2$ 5 times, We can get $A + 5B = 6517-1 = 6516$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$A = 6516 - 5B$, Obviously, we have $1303$ different pairs.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 2: We used Spell $1$ 2 times and Spell $2$ 4 times, We can get $2A + 4B = 6517-1 = 6516$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$A = 3258 - 2B$, Obviously, we have $1628$ different pairs.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 3: We used Spell $1$ 3 times and Spell $2$ 3 times, We can get $3A + 3B = 6517-1 = 6516$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$A = 2172 - B$, Obviously, we have $2171$ different pairs.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 4: We used Spell $1$ 4 times and Spell $2$ 2 times. It is equivalent with Case 2. We have $1628$ different pairs.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 5: We used Spell $1$ 5 times and Spell $2$ 1 time. It is equivalent with Case 1. We have $1303$ different pairs.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nHowever, we need to remove the pairs in more than one cases.\nWhen $a \\neq b$, there are no pairs in more than one cases.\nWhen $a = b = 1086$, there is a pair in all five cases, because different coefficients can't change the final value on the blackboard.\u0026lt;\/p\u0026gt;\n\nThus, there are $1303+1628+2171+1628+1303-1\\times 4=\\boxed{8029}$ pairs.\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9268053}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_accdef91-0e13-454f-bd7e-87cc3a902348","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_02b421bc-19a5-4075-9b60-63c170db826d","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"532c715b-62ba-4ecb-8002-94ca29f591fd","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"e9f069a4-db51-468c-83c8-e44063f96f4b","items":[{"type":"BlockComponentItem","id":"3802684e-dcb0-466a-b580-8f0a00ee3624","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"54103e1a-dd1e-46c4-99f0-690b26139f02","items":[{"type":"HtmlComponent","id":26989943,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page11042026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270811}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"11 Apr 2026","uid":"c43bd40a-01e9-4ee4-90db-66c8ca0a1780","path":"\/11apr2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_8a88824c-ca15-4a49-9cbd-109df5a5d780","sections":[{"type":"Slide","id":"f_320deb80-1f74-41a9-957f-8c7f4c66a7f1","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_83b99538-3a5e-4c79-9b4f-c195b697523c","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_24101386-6889-47eb-a94a-f7e2fd5d087d","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_3f2b53a3-c6cd-4339-b110-e8afacb575b7","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_81d7708f-eaec-4d77-ba8b-b890573a9df3","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_0199c5f4-00a9-468c-bee0-37660555128a","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_58b8f221-15f5-4011-a147-77c374fcf935","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_e9ed2b43-c064-4023-a999-466bb35290a7","defaultValue":null,"items":[{"type":"HtmlComponent","id":26989934,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

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    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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  • —— Hello

    Let's Solve Problems!

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  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;Zoom bought two decks of poker cards and discarded the Jokers. He picked 26 cards from the rest 104 cards. If the probability that there are at least 6 bombs is $P$, Find the nearest integer from $10^{10}P$.\n(Notes: A bomb is a combination with at least 4 cards with same rank, but 8 same cards are considered only ONE bomb instead of two bombs. For example, \"AAAKKKKKK777777776666\" includes three bombs, even if there are two \"7777\")\u0026lt;\/p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;The total number of ways to choose cards is $\\binom{104}{26} \\approx 2.2551015 \\cdot 10^{24}$, and then we can do casework:\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;Case 1($6 \\cdot 4 + 2$): We have $\\binom{13}{6} \\cdot (\\binom{8}{4})^6 \\cdot \\binom{56}{2} \\approx 3.10903953 \\cdot 10^{17}$ ways\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Case 2($1\\cdot 5 + 5 \\cdot 4 + 1$): We have $\\frac{P(13,6)}{120} \\cdot (\\binom{8}{5}) \\cdot (\\binom{8}{4})^5 \\cdot \\binom{56}{1} \\approx 5.42668719 \\cdot 10^{16}$ ways\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Case 3($1\\cdot 6 + 5 \\cdot 4$): We have $\\frac{P(13,6)}{120} \\cdot (\\binom{8}{6}) \\cdot (\\binom{8}{4})^5 \\approx 4.84525642 \\cdot 10^{14}$ ways\u0026lt;\/p\u0026gt;\n \n\u0026lt;p\u0026gt;Case 4($2\\cdot 5 + 4 \\cdot 4$): We have $\\frac{P(13,6)}{24 \\cdot 2} \\cdot (\\binom{8}{5})^2 \\cdot (\\binom{8}{4})^4 \\approx 1.93810257 \\cdot 10^{15}$ ways\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;To sum up, we have approximately $3.6759345387 \\cdot 10^{17}$ ways to choose cards. Thus, the probability is $\\frac{3.6759345387 \\cdot 10^{17}}{2.2551015 \\cdot 10^{24}}$, and we have $10^{10}P \\approx 1630.053 \\approx \\boxed{1630}$\u0026lt;\/p\u0026gt;\n\n\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270809}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_979f1b2e-5fde-4128-a01f-d7219fbad647","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_f5d976fe-ac49-440d-8e16-da0aadfc91fd","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"f0697b6b-6ae5-4e96-a101-785439aa37a1","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"7119ae5e-3f4f-48f1-a1c9-d48d4649f1b2","items":[{"type":"BlockComponentItem","id":"66e34338-9398-47c4-82d5-038ebe822e6b","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"e1b90c39-6ae6-407e-a04a-8d6b2b96d4cc","items":[{"type":"HtmlComponent","id":26989940,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page18042026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270812}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"18 Apr 2026","uid":"6ecf13b1-3dc0-4672-a669-ffc39944a53e","path":"\/18apr2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_058e8602-fd2a-428d-ad0c-f87f41d3dcfe","sections":[{"type":"Slide","id":"8f82740e-1ce0-46ba-82bd-815ea1fd01b5","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"12f046d2-840b-48b2-9671-931f05278c65","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"9626e19d-1d23-4310-82c6-ccf387dc60a7","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"224f494f-83ed-4594-b448-4f9937999dec","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"1821eb51-18b7-4466-b150-7c1b7ae85cef","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"095be0e7-2258-4e39-bacd-7fc614e0efee","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"734e203c-9a21-4f6c-9fb7-7f7d5abdba80","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"d6aa2c48-f457-4701-8c3a-4043a16f8f02","defaultValue":null,"items":[{"type":"HtmlComponent","id":27036966,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n \n \n\n \u0026lt;p\u0026gt;Pick $n(n\u0026gt;1)$ integers between 2 and 20, inclusive. Find the number of ways that any two integers of them are coprime.\u0026lt;\/p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;We can group these 19 numbers\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 2(only divisible by 2): 2 4 8 16\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 3(only divisible by 3): 3 9\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 5(only divisible by 5): 5\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 7(only divisible by 7): 7\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 2,3(only divisible by 2 and 3): 6 12 18\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 2,5(only divisible by 2 and 5): 10 20\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 2,7(only divisible by 2 and 7): 14\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group 3,5(only divisible by 3 and 5): 15\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Group X(not divisible by 2, 3, 5 or 7): 11 13 17 19\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;First, we can observe these numbers has following properties:\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;1. The number of numbers besides Group X is at most 4.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;2. All numbers in Group X are coprime with any number else\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;We can do casework base the number of numbers chosen from the other 8 groups\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 4: $Ways = 4 \\cdot 2 \\cdot 1 \\cdot 1 \\cdot 2^4 = 128$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 3: \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt; Case 3.1(Choose 2 numbers from group 2 3 5 7): \n$Ways = 3 \\cdot (1 \\cdot 1) + 2 \\cdot (2 \\cdot 1) + 1 \\cdot (2 \\cdot 1) + 1 \\cdot (4 \\cdot 1) = 13$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt; Case 3.2(Choose all 3 numbers from group 2 3 5 7): \n$Ways = 4 \\cdot 2 \\cdot 1 + 4 \\cdot 2 \\cdot 1 + 4 \\cdot 1 \\cdot 1 + 2 \\cdot 1 \\cdot 1 = 22$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;So, in case 3, we have $(13+22) \\cdot 2^4 = 560$ ways\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Case 2: \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt; Case 2.1(Choose all 2 numbers from group 2 3 5 7): \n$Ways = 4 \\cdot 2 + 4 \\cdot 1 + 4 \\cdot 1 + 2 \\cdot 1 + 2\\cdot 1 + 1 \\cdot 1 = 21$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt; Case 2.2(Choose 1 number from group 2 3 5 7): \n$Ways = 4 \\cdot 1 + 2 \\cdot (2+1) + 1 \\cdot (3+1) + 1 \\cdot (3+2+1) = 20$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt; Case 2.3(Choose 0 numbers from group 2 3 5 7): \n$Ways = 1$(Choose 14 and 15)\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;So, in case 2, we have $(21+20+1) \\cdot 2^4 = 672$ ways\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Case 1: Any of 15 numbers can be selected, so we have $15 \\cdot (2^4-1) = 225$ ways(We have to choose at least 1 number in Group X, because we need at least 2 numbers in total.)\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Case 0: We have $1 \\cdot (2^4-4-1) = 11$ ways(We have to choose at least 2 number in Group X). \u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Thus, we have $128+560+672+225+11=\\boxed{1596}$ ways\u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270809}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"2936f48f-8294-45bb-9b0d-5d0a80299318","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"00f4f98a-bbe7-4731-a2f7-60c7ea296523","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"268b2220-50d4-4ab9-8bd1-e939527cfcde","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"c12c65d3-00d3-40f3-9e6c-efdcf88f91bf","items":[{"type":"BlockComponentItem","id":"202dff5e-5a15-466d-8708-89451694a7ef","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"a5a9678d-5bd5-4259-81be-e3cfae0715af","items":[{"type":"HtmlComponent","id":27036967,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page25042026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270812}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"25 Apr 2026","uid":"8d51ab1c-417e-4619-8d49-bbbdafa89586","path":"\/25apr2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_9b2e39f9-fd22-4ea7-9668-91682e29167e","sections":[{"type":"Slide","id":"da23dccf-3259-4e5c-b8d8-2eca709be0c0","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"29fb96d8-adfc-4e83-b73b-c0d4b82c9a70","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"179ef546-6c93-48a4-99ca-a9df89e66c05","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"66fa67bd-4ef4-477a-b7f8-ba0407a298b6","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"17429f5f-7a63-4d03-a1e5-36c938170e98","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"84d8d8e4-c0e9-4cb4-b94f-c162910bb9c1","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"a7e6e27f-2622-4687-8b91-234d258876de","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"7315646d-a418-4e93-924c-d1143580a622","defaultValue":null,"items":[{"type":"HtmlComponent","id":27078477,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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  • This Week's Problem

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    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

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  • —— Hello

    Let's Solve Problems!

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  • This Week's Problem

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    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n \n \n\n \u0026lt;p\u0026gt; How many numbers are divisible by $2475$ in all permutations of $123456789$? \u0026lt;\/p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;p\u0026gt;1. Factor 2475: $2475 = 3^2 \\cdot 5^2 \\cdot 11 $\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;2. Consider it is divisible by 9\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Let $\\overline{ABCDEFGHI}$ be the number, which is one of all permutations of $123456789$. We can find $A+B+C+D+E+F+G+H+I=45$, thus it must be divisible by 9\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;3. Consider it is divisible by 25\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Obviously, $H = 2$ or $7$ and $I = 5$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;4. Consider it is divisible by 11\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Let $S_1$ be $A+C+E+G+I$ and $S_2$ be $B+D+F+H$, and $S_1+S_2 = 45$ and $|S_1-S_2| = 11k$ \u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;We can find these valid pairs $(S_1,S_2)$:\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$(28,17)$ and $(17,28)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 1: $(S_1,S_2)$ = $(28,17)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$\\begin{equation*}\n\\begin{cases}\nA+C+E+G = 23 \\\\\nB+D+F+H = 17\n\\end{cases}\n\\end{equation*}$\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;Without Loss of Generality, we assume $B\\lt D \\lt F$ and $A\\lt C\\lt E\\lt G$\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;If $H=7$, we can find $B+D+F=10$. There is only $(B,D,F) = (1,3,6)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;If $H=2$, we can find $B+D+F=15$. These are $(B,D,F) = (1,6,8),(3,4,8)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Thus, in Case 1, we have $3 \\cdot 3! \\cdot 4! = 432$ ways\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Case 2: $(S_1,S_2)$ = $(17,28)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$\\begin{equation*}\n\\begin{cases}\nA+C+E+G = 12 \\\\\nB+D+F+H = 28\n\\end{cases}\n\\end{equation*}$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Without Loss of Generality, we assume $B\\lt D \\lt F$ and $A\\lt C\\lt E\\lt G$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Because $I = 5$, obviously, there are only $(A,C,E,G)=(1,2,3,6)$, $H = 7$, $(B,D,F) = (4,8,9)$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Thus, in Case 2, we have $1 \\cdot 3! \\cdot 4! = 144$ ways\u0026lt;\/p\u0026gt;\n \n\u0026lt;p\u0026gt;Thus, there are $432+144=\\boxed{576}$ ways\u0026lt;\/p\u0026gt; \n\n \n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270809}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"ff6c93af-37f4-4729-a532-5e672d00c7ed","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"c1bdb92f-6022-4da6-9542-ef2f4320e074","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"cd364fc7-18c6-4f1c-acf2-de39087f6b05","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"4a4f812d-7d9e-416e-86b0-9a2eedc4c488","items":[{"type":"BlockComponentItem","id":"691bb7c9-01b0-4bdc-8ecb-0005e9866dbe","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"3aad5138-909a-4e99-944c-f80f3a2521da","items":[{"type":"HtmlComponent","id":27078476,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page02052026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9270812}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"2 May 2026","uid":"b620db93-c265-4ecb-8bbc-ab3a81742a63","path":"\/2may2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_02444e74-d2ac-4a69-b8f0-5d038181c29c","sections":[{"type":"Slide","id":"f_71a2ebaf-9641-4591-b5cf-44ec52d2767c","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_1e439ecc-cd43-4129-8787-79ed6ff1c70b","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_043b3b6d-7aba-4f9c-8067-456a6ea93ace","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_bb191009-71b4-49d3-8c00-566dac7c00f3","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_f564955f-3627-4ee5-a87a-2a5807418d0c","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_d5a91aa6-880b-419f-8d99-9b1d7468024f","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_8f9a80bd-8f10-479f-b017-cbe409fd5cd7","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_25716ee6-ce97-46fd-8dce-a40542968744","defaultValue":null,"items":[{"type":"HtmlComponent","id":27121677,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;In $\\triangle ABC$, let $A, B$, and $C$ be the interior angles. It is given that:$\n\\begin{equation*}\n\\sin A(1+\\cos B) + \\sin B(1+\\cos A) = \\frac{161}{65}\n\\end{equation*}$\nIf the circumradius and the inradius of the triangle is $65$ and $23$, determine the area of $\\triangle ABC$. \u0026lt;\/p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;$\\sin A(1+\\cos B) + \\sin B(1+\\cos A) = \\sin A + \\sin B + \\sin A \\cos B + \\cos A \\sin B = \\sin A + \\sin B + \\sin C = \\frac{161}{65}$ \u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;By Sine Law, which is $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2R$ \u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;$\\frac{a+b+c}{\\sin A+\\sin B+\\sin C}= 2R$ \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$\\frac{a+b+c}{\\frac{161}{65}}= 2 \\cdot 65$ \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$a+b+c= 322$ \u0026lt;\/p\u0026gt; \n\u0026lt;p\u0026gt;Let the incenter of the triangle be point $I$, join $IA$, $IB$ and $IC$.\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$S_{IAB} = \\frac{rc}{2},S_{IBC} = \\frac{ra}{2},S_{IAC} = \\frac{rb}{2}$\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;$S_{ABC} = S_{IAB}+S_{IBC}+S_{IAC} = \\frac{r(a+b+c)}{2} = \\frac{r(a+b+c)}{2} = \\frac{r(a+b+c)}{2} = \\frac{322 \\cdot 23}{2} = \\boxed{3703}$ \u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9282680}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_8d74adc0-c4ae-447c-9d6f-f064570726be","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_5182d007-06a0-4abc-9a7e-8520f0e8ab71","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"67a0d979-d3fd-47dc-9268-f91fd1ff88b4","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"2a40aab4-21b6-4eb9-90bd-67ecfe9a5a5f","items":[{"type":"BlockComponentItem","id":"c42fb652-1108-4769-beb4-235e1905d74a","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"0db9d291-0abb-461a-822d-454bd6c84bcf","items":[{"type":"HtmlComponent","id":27121881,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page09052026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9282741}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"9 May 2026","uid":"9cbfeb65-dcc2-4ea7-87d6-f30173c81ed4","path":"\/9may2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_27444594-6f64-47f0-a9b8-a9cf59cbe2da","sections":[{"type":"Slide","id":"f_7666412d-483e-4a95-9e1e-b65514aa76e5","defaultValue":true,"template_name":"html","components":{"slideSettings":{"type":"SlideSettings","id":"f_1ed2d022-12e4-4251-b5ca-fbeb95bbbfd4","defaultValue":true,"show_nav":true,"hidden_section":false,"name":"Embed an App","sync_key":null},"text1":{"type":"RichText","id":"f_facd9127-1756-48b2-aff0-43668aed23d6","defaultValue":false,"value":"","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":null,"lastLineTextAlign":null},"defaultDataProcessed":true},"text2":{"type":"RichText","id":"f_31d06408-803e-43e4-9ada-b0cc80e70b9c","defaultValue":false,"value":"","backupValue":null,"version":1,"lineAlignment":{"firstLineTextAlign":null,"lastLineTextAlign":null},"defaultDataProcessed":true},"background1":{"type":"Background","id":"f_29346701-9cba-4bd6-ac9f-6e5a0c0f5d51","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","videoUrl":"","videoHtml":""},"html1":{"type":"HtmlComponent","id":27167433,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

  • —— Hello

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  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

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  • —— Hello

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt; \"Xanthos and Balthazar both visit a coffee shop on Day $0$. For any subsequent Day $n$:If Balthazar did not visit the shop on Day $n$, then on Day $n+1$, the probability of Balthazar visiting is $\\frac{4}{5}$, and the probability of Xanthos visiting is $\\frac{9}{10}$.If Balthazar did visit the shop on Day $n$, then on Day $n+1$, the probability of Balthazar visiting is $\\frac{1}{5}$, and the probability of Xanthos visiting is $\\frac{1}{20}$. Special Rule: If Balthazar does not visit the shop for two consecutive days, the probability of Xanthos visiting on the following day becomes $\\frac{99}{100}$. Assuming Xanthos and Balthazar decide to visit independently each day based on these conditions, If the expected number of days until their next simultaneous visit (encounter) can be written as $\\frac{a}{b}$, where $gcd(a,b) = 1$. Find $a+3b$\"\u0026lt;\/p\u0026gt;\n \u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \u0026lt;p\u0026gt;We can find the expected number of days until Xanthos and Balthazar simultaneously visit the shop by tracking Balthazar's history.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Let Day $0$ be the starting day when both friends visit the coffee shop together.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Since Balthazar's behavior is independent of Xanthos, we can define three states based on how many consecutive days Balthazar has missed:\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;State 1 ($S_1$): Balthazar visited the shop today.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;State 2 ($S_2$): Balthazar missed today, but he visited yesterday (1 day missed consecutively).\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;State 3 ($S_3$): Balthazar missed both today and yesterday (2 or more days missed consecutively).\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Since both friends are at the shop on Day $0$, our system starts directly in State 1.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Now, let us determine the probability of a simultaneous encounter tomorrow from each state:\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;\u0026lt;b\u0026gt;From State 1:\u0026lt;\/b\u0026gt; Balthazar has a $\\frac{1}{5}$ chance to visit tomorrow, and Xanthos has a $\\frac{1}{20}$ chance.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;The encounter probability is: $\\frac{1}{5} \\times \\frac{1}{20} = \\frac{1}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If they do not meet but Balthazar still visits, we stay in State 1. The probability is: $\\frac{1}{5} - \\frac{1}{100} = \\frac{19}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If Balthazar does not visit, the system transitions to State 2. The probability is: $\\frac{4}{5} = \\frac{80}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;\u0026lt;b\u0026gt;From State 2:\u0026lt;\/b\u0026gt; Balthazar has a $\\frac{4}{5}$ chance to visit tomorrow, and Xanthos has a $\\frac{9}{10}$ chance.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;The encounter probability is: $\\frac{4}{5} \\times \\frac{9}{10} = \\frac{36}{50} = \\frac{72}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If they do not meet but Balthazar still visits, we return to State 1. The probability is: $\\frac{4}{5} - \\frac{36}{50} = \\frac{8}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If Balthazar does not visit, he has missed 2 days in a row, moving us to State 3. The probability is: $\\frac{1}{5} = \\frac{20}{100}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;\u0026lt;b\u0026gt;From State 3:\u0026lt;\/b\u0026gt; Balthazar has a $\\frac{4}{5}$ chance to visit, and Xanthos has a $\\frac{99}{100}$ chance due to the special rule.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;The encounter probability is: $\\frac{4}{5} \\times \\frac{99}{100} = \\frac{396}{500}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If they do not meet but Balthazar still visits, we return to State 1. The probability is: $\\frac{4}{5} - \\frac{396}{500} = \\frac{4}{500}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;If Balthazar stays home again, we remain in State 3. The probability is: $\\frac{1}{5} = \\frac{100}{500}$.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Let $E_1, E_2, E_3$ be the expected additional days until their next encounter from each state respectively.\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Each day that passes adds $1$ to the total count. We set up our system of equations using total expectation:\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Equation 1: $E_1 = 1 + \\frac{19}{100}E_1 + \\frac{4}{5}E_2$\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Equation 2: $E_2 = 1 + \\frac{2}{25}E_1 + \\frac{1}{5}E_3$\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Equation 3: $E_3 = 1 + \\frac{1}{125}E_1 + \\frac{1}{5}E_3$\u0026lt;\/p\u0026gt;\n \n \u0026lt;p\u0026gt;Solving for $E_1$ gives the total expected number of days: $E_1 = \\frac{5000}{1861}$\u0026lt;\/p\u0026gt;\n \n \n \u0026lt;p\u0026gt;Thus, we have $a = 5000$ and $b = 1861$.\u0026lt;\/p\u0026gt;\n\n \n \u0026lt;p\u0026gt;$a + 3b = 5000 + 3 \\cdot 1861 = \\boxed{10583}$\u0026lt;\/p\u0026gt;\n \n\n \n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9286146}"}}},{"type":"Slide","id":"f_85fba786-00ee-49c3-95d5-3ddf88d5d9fc","defaultValue":true,"template_name":"html","components":{"slideSettings":{"type":"SlideSettings","id":"f_55f282fd-e79e-44d6-ba44-8264acb037cd","defaultValue":true,"show_nav":true,"hidden_section":false,"name":"Embed an App","sync_key":null}}}],"title":"16 May 2026","uid":"bbb38d3e-a5dc-402a-a555-96293907f28f","path":"\/16may2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_accfa930-101e-4221-a475-0e94a9201253","sections":[{"type":"Slide","id":"f_c4aff386-1e51-4af6-aa8f-3c9afd1d0d27","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_0bccf08d-1113-40fd-9242-578e8c38c5da","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_1363d009-6f03-4ad7-b706-c8a055a245dc","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_dafd80c0-d886-435c-87b0-5f0a262c669a","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_9b4bd3cd-b34f-4ce3-a914-597167cff8b8","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_337e225a-fc35-4ba4-9391-a93a3df3497e","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_db454fa4-9951-45fa-9bc4-dc8396e0f8d8","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_a0f26609-a1d7-4ade-9c4a-0a75adfe08d4","defaultValue":null,"items":[{"type":"HtmlComponent","id":27204087,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '

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, right: '

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    Let's Solve Problems!

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  • This Week's Problem

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    You can click on the history page to check it.

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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;Find the last three digits of $2023^{2024^{2025}}$\u0026lt;\/p\u0026gt;\n \u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;By Euler's Theorem, $2023^{2024^{2025}} \\equiv 23^{24^{2025} \\bmod 400} \\pmod{1000}$\u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;By bashing, we can find:\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;$24^{1} \\bmod 400 = 24$\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;$24^{2} \\bmod 400 = 176$\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;$24^{3} \\bmod 400 = 224$\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;$24^{4} \\bmod 400 = 176$\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;...\u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;$24^{2025} \\bmod 400 = 224$\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;$2023^{2024^{2025}} \\equiv 23^{224} \\equiv 41 \\pmod{1000}$(Exponentiation by Squaring)\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;Thus, the last 3 digits are $\\boxed{041}$\u0026lt;\/p\u0026gt;\n\n\n \n \n \n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9288482}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_4e706460-d71d-4139-a5d7-3facbc9b03f4","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_25a8abbf-a057-46a0-b8c7-8ec553ec0a94","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"bc6a11b6-ff58-429d-b522-fd639b598dd0","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"77f1c4ce-e7d3-4b84-85ed-6e52b9da8bcc","items":[{"type":"BlockComponentItem","id":"91c34c5e-ef1c-4bbc-9816-37b3d12f4043","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f23638fa-f4fb-4d7c-bebf-1725b318e49d","items":[{"type":"HtmlComponent","id":27204088,"defaultValue":false,"value":"\u0026lt;div 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  • —— Hello

    Let's Solve Problems!

    We publish a new problem every Saturday! Join us!

  • This Week's Problem

    The solution will be released next week(6 Jun 2026).

    You can click on the history page to check it.

    Reminder: The answer of our problem may be very big, so do not try answer one by one. It does not work.

    The answer must be an integer from -1 googol to 1 googol(10^100).

    Our problem and the answer of last week problem is released every Saturday.

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