To find the remainder of 738mod\u2009\u20096607^{38} \\mod 660738mod660, we can use the Chinese Remainder Theorem<\/strong>. Since 660=22\u00d73\u00d75\u00d711660 = 2^2 \\times 3 \\times 5 \\times 11660=22\u00d73\u00d75\u00d711, we first need to compute 738mod\u2009\u200947^{38} \\mod 4738mod4, 738mod\u2009\u200937^{38} \\mod 3738mod3, 738mod\u2009\u200957^{38} \\mod 5738mod5, and 738mod\u2009\u2009117^{38} \\mod 11738mod11, then combine the results to find 738mod\u2009\u20096607^{38} \\mod 660738mod660.<\/p>\n\n\n\n Since 7\u2261\u22121mod\u2009\u200947 \\equiv -1 \\mod 47\u2261\u22121mod4, we have:738\u2261(\u22121)38\u22611mod\u2009\u20094.7^{38} \\equiv (-1)^{38} \\equiv 1 \\mod 4.738\u2261(\u22121)38\u22611mod4.<\/p>\n\n\n\n Since 7\u22611mod\u2009\u200937 \\equiv 1 \\mod 37\u22611mod3, we have:738\u2261138\u22611mod\u2009\u20093.7^{38} \\equiv 1^{38} \\equiv 1 \\mod 3.738\u2261138\u22611mod3.<\/p>\n\n\n\n We use Fermat\u2019s Little Theorem<\/strong>: since 555 is prime, 74\u22611mod\u2009\u200957^{4} \\equiv 1 \\mod 574\u22611mod5. We can reduce the exponent 38mod\u2009\u2009438 \\mod 438mod4, which gives:38\u00f74=9 remainder 2.38 \\div 4 = 9 \\text{ remainder } 2.38\u00f74=9 remainder 2.<\/p>\n\n\n\n Thus, 738\u226172mod\u2009\u200957^{38} \\equiv 7^2 \\mod 5738\u226172mod5. Now, 72=497^2 = 4972=49, and 49mod\u2009\u20095=449 \\mod 5 = 449mod5=4. Therefore:738\u22614mod\u2009\u20095.7^{38} \\equiv 4 \\mod 5.738\u22614mod5.<\/p>\n\n\n\n Again using Fermat\u2019s Little Theorem<\/strong> for p=11p = 11p=11, we know 710\u22611mod\u2009\u2009117^{10} \\equiv 1 \\mod 11710\u22611mod11. We reduce 38mod\u2009\u20091038 \\mod 1038mod10, which gives:38\u00f710=3 remainder 8.38 \\div 10 = 3 \\text{ remainder } 8.38\u00f710=3 remainder 8.<\/p>\n\n\n\n Thus, 738\u226178mod\u2009\u2009117^{38} \\equiv 7^8 \\mod 11738\u226178mod11. Now, calculating powers of 777 modulo 111111:72=49\u22615mod\u2009\u200911,74=52=25\u22613mod\u2009\u200911,78=32=9mod\u2009\u200911.7^2 = 49 \\equiv 5 \\mod 11, \\quad 7^4 = 5^2 = 25 \\equiv 3 \\mod 11, \\quad 7^8 = 3^2 = 9 \\mod 11.72=49\u22615mod11,74=52=25\u22613mod11,78=32=9mod11.<\/p>\n\n\n\n So, 738\u22619mod\u2009\u2009117^{38} \\equiv 9 \\mod 11738\u22619mod11.<\/p>\n\n\n\n We now have the system of congruences:x\u22611mod\u2009\u20094,x\u22611mod\u2009\u20093,x\u22614mod\u2009\u20095,x\u22619mod\u2009\u200911.\\begin{aligned} x &\\equiv 1 \\mod 4, \\\\ x &\\equiv 1 \\mod 3, \\\\ x &\\equiv 4 \\mod 5, \\\\ x &\\equiv 9 \\mod 11. \\end{aligned}xxxx\u200b\u22611mod4,\u22611mod3,\u22614mod5,\u22619mod11.\u200b<\/p>\n\n\n\n We can start by solving the system for xmod\u2009\u2009660x \\mod 660xmod660. First, solve x\u22611mod\u2009\u200912x \\equiv 1 \\mod 12x\u22611mod12 (since 444 and 333 are relatively prime). Now, the system is:x\u22611mod\u2009\u200912,x\u22614mod\u2009\u20095,x\u22619mod\u2009\u200911.\\begin{aligned} x &\\equiv 1 \\mod 12, \\\\ x &\\equiv 4 \\mod 5, \\\\ x &\\equiv 9 \\mod 11. \\end{aligned}xxx\u200b\u22611mod12,\u22614mod5,\u22619mod11.\u200b<\/p>\n\n\n\n Next, solve x\u22611mod\u2009\u200912x \\equiv 1 \\mod 12x\u22611mod12 and x\u22614mod\u2009\u20095x \\equiv 4 \\mod 5x\u22614mod5. Using the method of successive substitution, we find that x\u226113mod\u2009\u200960x \\equiv 13 \\mod 60x\u226113mod60. Finally, solve the system:x\u226113mod\u2009\u200960,x\u22619mod\u2009\u200911.x \\equiv 13 \\mod 60, \\quad x \\equiv 9 \\mod 11.x\u226113mod60,x\u22619mod11.<\/p>\n\n\n\n Using the method of successive substitution again, we find:x\u226173mod\u2009\u2009660.x \\equiv 73 \\mod 660.x\u226173mod660.<\/p>\n\n\n\n The remainder when 7387^{38}738 is divided by 660 is 73<\/strong>.<\/p>\n\n\n\n find the remainder of 7^38 when divided by 660 The Correct Answer and Explanation is: To find the remainder of 738mod\u2009\u20096607^{38} \\mod 660738mod660, we can use the Chinese Remainder Theorem. Since 660=22\u00d73\u00d75\u00d711660 = 2^2 \\times 3 \\times 5 \\times 11660=22\u00d73\u00d75\u00d711, we first need to compute 738mod\u2009\u200947^{38} \\mod 4738mod4, 738mod\u2009\u200937^{38} \\mod 3738mod3, 738mod\u2009\u200957^{38} \\mod 5738mod5, and […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-264036","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/264036","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=264036"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/264036\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=264036"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=264036"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=264036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate 738mod\u2009\u200947^{38} \\mod 4738mod4<\/h3>\n\n\n\n
Step 2: Calculate 738mod\u2009\u200937^{38} \\mod 3738mod3<\/h3>\n\n\n\n
Step 3: Calculate 738mod\u2009\u200957^{38} \\mod 5738mod5<\/h3>\n\n\n\n
Step 4: Calculate 738mod\u2009\u2009117^{38} \\mod 11738mod11<\/h3>\n\n\n\n
Step 5: Solve using the Chinese Remainder Theorem<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"