{"id":90250,"date":"2025-06-01T10:23:54","date_gmt":"2025-06-01T10:23:54","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90250"},"modified":"2025-06-01T10:23:54","modified_gmt":"2025-06-01T10:23:54","slug":"which-one-of-the-following-is-the-remainder-when-1020-is-divided-by","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/","title":{"rendered":"Which one of the following is the remainder when 10^20 is divided by"},"content":{"rendered":"

Which one of the following is the remainder when 10^20 is divided by 7?<\/p>\n

[amp_mcq option1=”1″ option2=”2″ option3=”4″ option4=”6″ correct=”option2″]<\/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 remainder when 10^20 is divided by 7 is 2.
\n<\/section>\n
\n– We need to compute 10^20 mod 7.
\n– First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7).
\n– Then, 10^20 \u2261 3^20 (mod 7).
\n– Now, we compute powers of 3 modulo 7 to find a pattern (cycle):
\n – 3^1 \u2261 3 (mod 7)
\n – 3^2 \u2261 9 \u2261 2 (mod 7)
\n – 3^3 \u2261 3 * 2 = 6 (mod 7)
\n – 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7)
\n – 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7)
\n – 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7)
\n– The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1).
\n– To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6.
\n– 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2).
\n– So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7).
\n– Since 3^6 \u2261 1 (mod 7), we have:
\n – 3^20 \u2261 (1)^3 * 3^2 (mod 7)
\n – 3^20 \u2261 1 * 9 (mod 7)
\n – 3^20 \u2261 9 \u2261 2 (mod 7).
\n– The remainder is 2.
\n<\/section>\n
\nThis method uses modular arithmetic properties, specifically finding the cyclic nature of powers modulo a number. Fermat’s Little Theorem could also be applied here since 7 is a prime number and 3 is not divisible by 7. Fermat’s Little Theorem states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) \u2261 1 (mod p). Here, a=3, p=7. So, 3^(7-1) = 3^6 \u2261 1 (mod 7). This confirms our calculated cycle length. Then we proceed as shown.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Which one of the following is the remainder when 10^20 is divided by 7? [amp_mcq option1=”1″ option2=”2″ option3=”4″ option4=”6″ correct=”option2″] This question was previously asked in UPSC CAPF – 2018 Download PDFAttempt Online The remainder when 10^20 is divided by 7 is 2. – We need to compute 10^20 mod 7. – First, simplify the … <\/p>\n

Detailed SolutionWhich one of the following is the remainder when 10^20 is divided by<\/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-90250","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1114","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nWhich one of the following is the remainder when 10^20 is divided by<\/title>\n<meta name=\"description\" content=\"The remainder when 10^20 is divided by 7 is 2. - We need to compute 10^20 mod 7. - First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7). - Then, 10^20 \u2261 3^20 (mod 7). - Now, we compute powers of 3 modulo 7 to find a pattern (cycle): - 3^1 \u2261 3 (mod 7) - 3^2 \u2261 9 \u2261 2 (mod 7) - 3^3 \u2261 3 * 2 = 6 (mod 7) - 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7) - 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7) - 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7) - The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1). - To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6. - 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2). - So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7). - Since 3^6 \u2261 1 (mod 7), we have: - 3^20 \u2261 (1)^3 * 3^2 (mod 7) - 3^20 \u2261 1 * 9 (mod 7) - 3^20 \u2261 9 \u2261 2 (mod 7). - The remainder is 2.\" \/>\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\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Which one of the following is the remainder when 10^20 is divided by\" \/>\n<meta property=\"og:description\" content=\"The remainder when 10^20 is divided by 7 is 2. - We need to compute 10^20 mod 7. - First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7). - Then, 10^20 \u2261 3^20 (mod 7). - Now, we compute powers of 3 modulo 7 to find a pattern (cycle): - 3^1 \u2261 3 (mod 7) - 3^2 \u2261 9 \u2261 2 (mod 7) - 3^3 \u2261 3 * 2 = 6 (mod 7) - 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7) - 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7) - 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7) - The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1). - To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6. - 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2). - So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7). - Since 3^6 \u2261 1 (mod 7), we have: - 3^20 \u2261 (1)^3 * 3^2 (mod 7) - 3^20 \u2261 1 * 9 (mod 7) - 3^20 \u2261 9 \u2261 2 (mod 7). - The remainder is 2.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:23:54+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":"Which one of the following is the remainder when 10^20 is divided by","description":"The remainder when 10^20 is divided by 7 is 2. - We need to compute 10^20 mod 7. - First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7). - Then, 10^20 \u2261 3^20 (mod 7). - Now, we compute powers of 3 modulo 7 to find a pattern (cycle): - 3^1 \u2261 3 (mod 7) - 3^2 \u2261 9 \u2261 2 (mod 7) - 3^3 \u2261 3 * 2 = 6 (mod 7) - 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7) - 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7) - 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7) - The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1). - To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6. - 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2). - So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7). - Since 3^6 \u2261 1 (mod 7), we have: - 3^20 \u2261 (1)^3 * 3^2 (mod 7) - 3^20 \u2261 1 * 9 (mod 7) - 3^20 \u2261 9 \u2261 2 (mod 7). - The remainder is 2.","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\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/","og_locale":"en_US","og_type":"article","og_title":"Which one of the following is the remainder when 10^20 is divided by","og_description":"The remainder when 10^20 is divided by 7 is 2. - We need to compute 10^20 mod 7. - First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7). - Then, 10^20 \u2261 3^20 (mod 7). - Now, we compute powers of 3 modulo 7 to find a pattern (cycle): - 3^1 \u2261 3 (mod 7) - 3^2 \u2261 9 \u2261 2 (mod 7) - 3^3 \u2261 3 * 2 = 6 (mod 7) - 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7) - 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7) - 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7) - The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1). - To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6. - 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2). - So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7). - Since 3^6 \u2261 1 (mod 7), we have: - 3^20 \u2261 (1)^3 * 3^2 (mod 7) - 3^20 \u2261 1 * 9 (mod 7) - 3^20 \u2261 9 \u2261 2 (mod 7). - The remainder is 2.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:23:54+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\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/","url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/","name":"Which one of the following is the remainder when 10^20 is divided by","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:23:54+00:00","dateModified":"2025-06-01T10:23:54+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The remainder when 10^20 is divided by 7 is 2. - We need to compute 10^20 mod 7. - First, simplify the base: 10 mod 7 = 3. So, 10 \u2261 3 (mod 7). - Then, 10^20 \u2261 3^20 (mod 7). - Now, we compute powers of 3 modulo 7 to find a pattern (cycle): - 3^1 \u2261 3 (mod 7) - 3^2 \u2261 9 \u2261 2 (mod 7) - 3^3 \u2261 3 * 2 = 6 (mod 7) - 3^4 \u2261 3 * 6 = 18 \u2261 4 (mod 7) - 3^5 \u2261 3 * 4 = 12 \u2261 5 (mod 7) - 3^6 \u2261 3 * 5 = 15 \u2261 1 (mod 7) - The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1). - To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6. - 20 \u00f7 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2). - So, 3^20 \u2261 3^(6*3 + 2) \u2261 (3^6)^3 * 3^2 (mod 7). - Since 3^6 \u2261 1 (mod 7), we have: - 3^20 \u2261 (1)^3 * 3^2 (mod 7) - 3^20 \u2261 1 * 9 (mod 7) - 3^20 \u2261 9 \u2261 2 (mod 7). - The remainder is 2.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-remainder-when-1020-is-divided-by\/#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":"Which one of the following is the remainder when 10^20 is divided by"}]},{"@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\/90250","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=90250"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90250\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90250"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90250"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90250"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}