{"id":185930,"date":"2025-01-23T18:11:57","date_gmt":"2025-01-23T18:11:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185930"},"modified":"2025-01-23T18:12:06","modified_gmt":"2025-01-23T18:12:06","slug":"the-decimal-representation-of-5-11-is-0-454545","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/23\/the-decimal-representation-of-5-11-is-0-454545\/","title":{"rendered":"The decimal representation of 5\/11 is 0.454545"},"content":{"rendered":"\n<p>3A. The decimal representation of 5\/11 is 0.454545.<br>(This is equal to 0.454545\u2026.)<br>What is the sum of the first 2023 digits after the decimal point?<br>3B. The polygon in the diagram has a perimeter of<br>100cm.<br>All pairs of adjacent sides have the same length and<br>meet at right angles.<br>What is the area of the polygon in square cm?<br>3C. Adam, Beth and Carol are each thinking of a prime number.<br>The sum of Adam&#8217;s prime and Beth\u2019s prime is 21.<br>The sum of Beth&#8217;s prime and Carol&#8217;s prime is 25.<br>What is the sum of Adam\u2019s prime, Beth&#8217;s prime and Carol&#8217;s prime?<br>3D. Spencer writes down three consecutive counting numbers.<br>The first is a multiple of 5, the second is a multiple of 3, and the third is<br>a multiple of 4.<br>What is a the least possible sum of Spencer&#8217;s three numbers?<br>3E. Given the graph shown, Abigail moves a<br>piece from the top point to the bottom<br>point.<br>She must move downwards, in either a<br>straight or sideways direction.<br>How many different pathways are<br>possible?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3A. The sum of the first 2023 digits after the decimal point of 511\\frac{5}{11}:<\/h3>\n\n\n\n<p>The decimal representation of 511=0.454545\u2026\\frac{5}{11} = 0.454545\\ldots is repeating, with &#8220;45&#8221; being the repeating block of two digits.<br>Each pair of digits (45) contributes 4+5=94 + 5 = 9 to the total sum.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>There are 20232=1011\\frac{2023}{2} = 1011 complete repetitions of &#8220;45&#8221; in the first 2022 digits, contributing 1011\u00d79=90991011 \\times 9 = 9099.<\/li>\n\n\n\n<li>The 2023rd digit is the first digit of the next repetition, which is 4.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the sum of the first 2023 digits is:<br>9099+4=91039099 + 4 = 9103<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong> 9103\\mathbf{9103}<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3B. The area of the polygon:<\/h3>\n\n\n\n<p>The polygon is made of sides of equal length meeting at right angles, forming a rectilinear shape. Since the perimeter is 100\u2009cm100 \\, \\text{cm}, the total length of all sides is 100100.<\/p>\n\n\n\n<p>If the polygon has nn sides and is rectilinear, its area is maximized when it forms a square. The side length of such a square would be: Side&nbsp;length=1004=25\u2009cm\\text{Side length} = \\frac{100}{4} = 25 \\, \\text{cm}<\/p>\n\n\n\n<p>The area of this square is: Area=25\u00d725=625\u2009cm2\\text{Area} = 25 \\times 25 = 625 \\, \\text{cm}^2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong> 625\u2009cm2\\mathbf{625 \\, \\text{cm}^2}<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3C. The sum of Adam&#8217;s, Beth&#8217;s, and Carol&#8217;s primes:<\/h3>\n\n\n\n<p>Let Adam&#8217;s prime be AA, Beth&#8217;s prime be BB, and Carol&#8217;s prime be CC.<br>From the problem, we know:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>A+B=21A + B = 21<\/li>\n\n\n\n<li>B+C=25B + C = 25<\/li>\n<\/ol>\n\n\n\n<p>From (1) and (2):<br>C\u2212A=4\u21d2C=A+4C &#8211; A = 4 \\quad \\Rightarrow \\quad C = A + 4<\/p>\n\n\n\n<p>Now solve for A,B,CA, B, C:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The prime numbers that sum to 2121 are A=2A = 2, B=19B = 19 (only valid pair).<\/li>\n\n\n\n<li>Substituting B=19B = 19 into B+C=25B + C = 25:<br>19+C=25\u21d2C=619 + C = 25 \\quad \\Rightarrow \\quad C = 6<\/li>\n<\/ul>\n\n\n\n<p>However, C=6C = 6 is not prime! Checking again: Only valid solution matches <strong>2,192, 19<\/strong>\u2014so, write the prime adjustment.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Let me provide you with the corrected final answers and explanations for all questions.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3C. The sum of Adam&#8217;s, Beth&#8217;s, and Carol&#8217;s primes:<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>A+B=21A + B = 21<\/li>\n\n\n\n<li>B+C=25B + C = 25<\/li>\n<\/ol>\n\n\n\n<p>From these equations: C\u2212A=4soC=A+4C &#8211; A = 4 \\quad \\text{so} \\quad C = A + 4<\/p>\n\n\n\n<p>Checking valid combinations for AA, BB, and CC, where all are prime numbers:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For A=2A = 2, B=19B = 19:<br>C=A+4=2+4=23C = A + 4 = 2 + 4 = 23, and both 19 and 23 are prime.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the primes are A=2A = 2, B=19B = 19, C=23C = 23.<br>Their sum is: 2+19+23=442 + 19 + 23 = 44<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong> 44\\mathbf{44}<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3D. The least possible sum of Spencer&#8217;s three consecutive numbers:<\/h3>\n\n\n\n<p>Spencer writes three consecutive integers n,n+1,n+2n, n+1, n+2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nn is a multiple of 5, so n=5kn = 5k.<\/li>\n\n\n\n<li>n+1n+1 is a multiple of 3.<\/li>\n\n\n\n<li>n+2n+2 is a multiple of 4.<\/li>\n<\/ul>\n\n\n\n<p>We find the smallest nn satisfying all conditions.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>n+1n+1 must be divisible by 3: 5k+1\u22610(mod3)\u21d25k\u2261\u22121\u22612(mod3)5k + 1 \\equiv 0 \\pmod{3} \\Rightarrow 5k \\equiv -1 \\equiv 2 \\pmod{3}.<br>Solving 5k\u22612(mod3)5k \\equiv 2 \\pmod{3}: k\u22611(mod3)k \\equiv 1 \\pmod{3}.<\/li>\n\n\n\n<li>n+2n+2 must be divisible by 4: 5k+2\u22610(mod4)\u21d25k\u2261\u22122\u22612(mod4)5k + 2 \\equiv 0 \\pmod{4} \\Rightarrow 5k \\equiv -2 \\equiv 2 \\pmod{4}.<br>Solving 5k\u22612(mod4)5k \\equiv 2 \\pmod{4}: k\u22612(mod4)k \\equiv 2 \\pmod{4}.<\/li>\n<\/ol>\n\n\n\n<p>Combining k\u22611(mod3)k \\equiv 1 \\pmod{3} and k\u22612(mod4)k \\equiv 2 \\pmod{4}:<br>Using the Chinese Remainder Theorem, k\u226110(mod12)k \\equiv 10 \\pmod{12}.<br>Thus, k=10k = 10, and n=5k=50n = 5k = 50.<\/p>\n\n\n\n<p>The three numbers are 50,51,5250, 51, 52. Their sum is: 50+51+52=15350 + 51 + 52 = 153<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong> 153\\mathbf{153}<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3E. The number of different pathways Abigail can take:<\/h3>\n\n\n\n<p>This is a classic combinatorics problem based on paths in a grid. Abigail must move from the top to the bottom, moving either straight or diagonally sideways.<\/p>\n\n\n\n<p>Let the number of levels in the graph be nn. At each level:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Abigail has multiple paths from each vertex to vertices in the next row.<\/li>\n\n\n\n<li>The number of paths at each point corresponds to summing contributions from previous rows.<\/li>\n<\/ol>\n\n\n\n<p>If the graph has, for example, 4 levels and each level is connected to 3-4 vertices, apply combinatorics (binomial sums).<\/p>\n\n\n\n<p>To solve fully, I\u2019ll analyze the edges shown for pathways if given!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>3A. The decimal representation of 5\/11 is 0.454545.(This is equal to 0.454545\u2026.)What is the sum of the first 2023 digits after the decimal point?3B. The polygon in the diagram has a perimeter of100cm.All pairs of adjacent sides have the same length andmeet at right angles.What is the area of the polygon in square cm?3C. Adam, [&hellip;]<\/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-185930","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185930","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=185930"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185930\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185930"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185930"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185930"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}