{"id":90245,"date":"2025-06-01T10:23:49","date_gmt":"2025-06-01T10:23:49","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90245"},"modified":"2025-06-01T10:23:49","modified_gmt":"2025-06-01T10:23:49","slug":"what-is-the-largest-value-for-n-natural-number-such-that-6n-divides","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/","title":{"rendered":"What is the largest value for n (natural number) such that 6^n divides"},"content":{"rendered":"

What is the largest value for n (natural number) such that 6^n divides the product of the first 100 natural numbers?<\/p>\n

[amp_mcq option1=”18″ option2=”33″ option3=”44″ option4=”48″ correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48.
\n<\/section>\n
\n– To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial.
\n– 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3.
\n– The highest power of a prime p dividing n! is given by Legendre’s formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + …
\n– For p=3 and n=100:
\n Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81)
\n = 33 + 11 + 3 + 1 = 48.
\n– For p=2 and n=100:
\n Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64)
\n = 50 + 25 + 12 + 6 + 3 + 1 = 97.
\n– So, 100! contains 2^97 * 3^48 * …
\n– For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48.
\n– The largest integer n satisfying both conditions is n = 48.
\n<\/section>\n
\nLegendre’s formula is a powerful tool for finding the exponent of a prime in the prime factorization of n!. This method works because every multiple of a prime p contributes at least one factor of p, every multiple of p^2 contributes an additional factor of p, and so on.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

What is the largest value for n (natural number) such that 6^n divides the product of the first 100 natural numbers? [amp_mcq option1=”18″ option2=”33″ option3=”44″ option4=”48″ correct=”option4″] This question was previously asked in UPSC CAPF – 2018 Download PDFAttempt Online The largest value for n such that 6^n divides the product of the first 100 … <\/p>\n

Detailed SolutionWhat is the largest value for n (natural number) such that 6^n divides<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1085],"tags":[1114,1102],"class_list":["post-90245","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1114","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nWhat is the largest value for n (natural number) such that 6^n divides<\/title>\n<meta name=\"description\" content=\"The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48. - To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial. - 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3. - The highest power of a prime p dividing n! is given by Legendre's formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + ... - For p=3 and n=100: Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81) = 33 + 11 + 3 + 1 = 48. - For p=2 and n=100: Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64) = 50 + 25 + 12 + 6 + 3 + 1 = 97. - So, 100! contains 2^97 * 3^48 * ... - For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48. - The largest integer n satisfying both conditions is n = 48.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is the largest value for n (natural number) such that 6^n divides\" \/>\n<meta property=\"og:description\" content=\"The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48. - To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial. - 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3. - The highest power of a prime p dividing n! is given by Legendre's formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + ... - For p=3 and n=100: Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81) = 33 + 11 + 3 + 1 = 48. - For p=2 and n=100: Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64) = 50 + 25 + 12 + 6 + 3 + 1 = 97. - So, 100! contains 2^97 * 3^48 * ... - For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48. - The largest integer n satisfying both conditions is n = 48.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:23:49+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"What is the largest value for n (natural number) such that 6^n divides","description":"The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48. - To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial. - 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3. - The highest power of a prime p dividing n! is given by Legendre's formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + ... - For p=3 and n=100: Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81) = 33 + 11 + 3 + 1 = 48. - For p=2 and n=100: Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64) = 50 + 25 + 12 + 6 + 3 + 1 = 97. - So, 100! contains 2^97 * 3^48 * ... - For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48. - The largest integer n satisfying both conditions is n = 48.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/","og_locale":"en_US","og_type":"article","og_title":"What is the largest value for n (natural number) such that 6^n divides","og_description":"The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48. - To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial. - 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3. - The highest power of a prime p dividing n! is given by Legendre's formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + ... - For p=3 and n=100: Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81) = 33 + 11 + 3 + 1 = 48. - For p=2 and n=100: Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64) = 50 + 25 + 12 + 6 + 3 + 1 = 97. - So, 100! contains 2^97 * 3^48 * ... - For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48. - The largest integer n satisfying both conditions is n = 48.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:23:49+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/","url":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/","name":"What is the largest value for n (natural number) such that 6^n divides","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:23:49+00:00","dateModified":"2025-06-01T10:23:49+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48. - To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial. - 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3. - The highest power of a prime p dividing n! is given by Legendre's formula: floor(n\/p) + floor(n\/p^2) + floor(n\/p^3) + ... - For p=3 and n=100: Power of 3 in 100! = floor(100\/3) + floor(100\/9) + floor(100\/27) + floor(100\/81) = 33 + 11 + 3 + 1 = 48. - For p=2 and n=100: Power of 2 in 100! = floor(100\/2) + floor(100\/4) + floor(100\/8) + floor(100\/16) + floor(100\/32) + floor(100\/64) = 50 + 25 + 12 + 6 + 3 + 1 = 97. - So, 100! contains 2^97 * 3^48 * ... - For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n \u2264 97 and n \u2264 48. - The largest integer n satisfying both conditions is n = 48.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-largest-value-for-n-natural-number-such-that-6n-divides\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"What is the largest value for n (natural number) such that 6^n divides"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90245","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=90245"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90245\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}