{"id":90966,"date":"2025-06-01T10:42:08","date_gmt":"2025-06-01T10:42:08","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90966"},"modified":"2025-06-01T10:42:08","modified_gmt":"2025-06-01T10:42:08","slug":"the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/","title":{"rendered":"The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + … + (1 \u00d7 2 \u00d7 3 \u00d7 …"},"content":{"rendered":"

The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + … + (1 \u00d7 2 \u00d7 3 \u00d7 … \u00d7 500) is divided by 8, is<\/p>\n

[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is A) 1.
\n<\/section>\n
\nWe need to find the remainder of the sum $S = 1! + 2! + 3! + … + 500!$ when divided by 8.
\nLet’s compute the first few factorials modulo 8:
\n$1! = 1 \\equiv 1 \\pmod 8$
\n$2! = 2 \\equiv 2 \\pmod 8$
\n$3! = 6 \\equiv 6 \\pmod 8$
\n$4! = 24 = 3 \\times 8 \\equiv 0 \\pmod 8$
\n$5! = 5 \\times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! \\equiv 0 \\pmod 8$.
\nIn general, for any integer $n \\ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n \\ge 4$.
\n<\/section>\n
\nSo, $n! \\pmod 8 = 0$ for $n \\ge 4$.
\nThe sum modulo 8 is:
\n$S \\pmod 8 = (1! + 2! + 3! + 4! + … + 500!) \\pmod 8$
\n$S \\pmod 8 = (1! \\pmod 8 + 2! \\pmod 8 + 3! \\pmod 8 + 4! \\pmod 8 + … + 500! \\pmod 8) \\pmod 8$
\n$S \\pmod 8 = (1 + 2 + 6 + 0 + 0 + … + 0) \\pmod 8$
\n$S \\pmod 8 = (1 + 2 + 6) \\pmod 8$
\n$S \\pmod 8 = 9 \\pmod 8$
\n$S \\pmod 8 = 1$.
\nThe remainder is 1.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + … + (1 \u00d7 2 \u00d7 3 \u00d7 … \u00d7 500) is divided by 8, is [amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option1″] This question was previously asked in UPSC CAPF – 2024 Download PDFAttempt Online The correct answer is A) … <\/p>\n

Detailed SolutionThe remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + … + (1 \u00d7 2 \u00d7 3 \u00d7 …<\/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":[1103,1102],"class_list":["post-90966","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1103","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nThe remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + ... + (1 \u00d7 2 \u00d7 3 \u00d7 ...<\/title>\n<meta name=\"description\" content=\"The correct answer is A) 1. We need to find the remainder of the sum $S = 1! + 2! + 3! + ... + 500!$ when divided by 8. Let's compute the first few factorials modulo 8: $1! = 1 equiv 1 pmod 8$ $2! = 2 equiv 2 pmod 8$ $3! = 6 equiv 6 pmod 8$ $4! = 24 = 3 times 8 equiv 0 pmod 8$ $5! = 5 times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! equiv 0 pmod 8$. In general, for any integer $n ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n ge 4$.\" \/>\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\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + ... + (1 \u00d7 2 \u00d7 3 \u00d7 ...\" \/>\n<meta property=\"og:description\" content=\"The correct answer is A) 1. We need to find the remainder of the sum $S = 1! + 2! + 3! + ... + 500!$ when divided by 8. Let's compute the first few factorials modulo 8: $1! = 1 equiv 1 pmod 8$ $2! = 2 equiv 2 pmod 8$ $3! = 6 equiv 6 pmod 8$ $4! = 24 = 3 times 8 equiv 0 pmod 8$ $5! = 5 times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! equiv 0 pmod 8$. In general, for any integer $n ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n ge 4$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:42:08+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":"The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + ... + (1 \u00d7 2 \u00d7 3 \u00d7 ...","description":"The correct answer is A) 1. We need to find the remainder of the sum $S = 1! + 2! + 3! + ... + 500!$ when divided by 8. Let's compute the first few factorials modulo 8: $1! = 1 equiv 1 pmod 8$ $2! = 2 equiv 2 pmod 8$ $3! = 6 equiv 6 pmod 8$ $4! = 24 = 3 times 8 equiv 0 pmod 8$ $5! = 5 times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! equiv 0 pmod 8$. In general, for any integer $n ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n ge 4$.","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\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/","og_locale":"en_US","og_type":"article","og_title":"The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + ... + (1 \u00d7 2 \u00d7 3 \u00d7 ...","og_description":"The correct answer is A) 1. We need to find the remainder of the sum $S = 1! + 2! + 3! + ... + 500!$ when divided by 8. Let's compute the first few factorials modulo 8: $1! = 1 equiv 1 pmod 8$ $2! = 2 equiv 2 pmod 8$ $3! = 6 equiv 6 pmod 8$ $4! = 24 = 3 times 8 equiv 0 pmod 8$ $5! = 5 times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! equiv 0 pmod 8$. In general, for any integer $n ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n ge 4$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:42:08+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\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-remainder-when-1-1-x-2-1-x-2-x-3-1-x-2-x-3-x\/","name":"The remainder, when 1 + (1 \u00d7 2) + (1 \u00d7 2 \u00d7 3) + ... + (1 \u00d7 2 \u00d7 3 \u00d7 ...","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:42:08+00:00","dateModified":"2025-06-01T10:42:08+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is A) 1. We need to find the remainder of the sum $S = 1! + 2! + 3! + ... + 500!$ when divided by 8. Let's compute the first few factorials modulo 8: $1! = 1 \\equiv 1 \\pmod 8$ $2! = 2 \\equiv 2 \\pmod 8$ $3! = 6 \\equiv 6 \\pmod 8$ $4! = 24 = 3 \\times 8 \\equiv 0 \\pmod 8$ $5! = 5 \\times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! \\equiv 0 \\pmod 8$. In general, for any integer $n \\ge 4$, $n!$ includes $4!$ as a factor. 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