{"id":213978,"date":"2025-05-13T08:41:21","date_gmt":"2025-05-13T08:41:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213978"},"modified":"2025-05-13T08:41:24","modified_gmt":"2025-05-13T08:41:24","slug":"on-a-table-there-are-1001-pennies-lined-up-in-a-row","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/on-a-table-there-are-1001-pennies-lined-up-in-a-row\/","title":{"rendered":"On a table, there are 1001 pennies lined up in a row"},"content":{"rendered":"\n<p>On a table, there are 1001 pennies lined up in a row. I then come along and replace every second penny with a nickel (starting with the second). After this I replace every third coin with a dime (starting with the third). Finally I replace every fourth coin with a quarter (starting with the fourth). How much money is on the table? Explain how you know using your knowledge of number theory and divisibility.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are given 1001 coins initially, all <strong>pennies<\/strong>, and then we make the following replacements:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Every <strong>2nd<\/strong> coin is replaced by a <strong>nickel (5\u00a2)<\/strong><\/li>\n\n\n\n<li>Every <strong>3rd<\/strong> coin is replaced by a <strong>dime (10\u00a2)<\/strong><\/li>\n\n\n\n<li>Every <strong>4th<\/strong> coin is replaced by a <strong>quarter (25\u00a2)<\/strong><\/li>\n<\/ol>\n\n\n\n<p>A coin can be replaced more than once, depending on its position. For example, position 12 is divisible by 2, 3, and 4 \u2014 so it is replaced three times.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-Step Plan:<\/h3>\n\n\n\n<p>Let\u2019s denote:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>P<\/strong> = Penny = 1\u00a2<\/li>\n\n\n\n<li><strong>N<\/strong> = Nickel = 5\u00a2<\/li>\n\n\n\n<li><strong>D<\/strong> = Dime = 10\u00a2<\/li>\n\n\n\n<li><strong>Q<\/strong> = Quarter = 25\u00a2<\/li>\n<\/ul>\n\n\n\n<p>Each position from 1 to 1001 starts with a penny. We go through and replace coins as described.<\/p>\n\n\n\n<p>We\u2019ll define a coin\u2019s final value based on the <em>last<\/em> replacement applied to it:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>First pass (every 2nd): replaces P \u2192 N<\/li>\n\n\n\n<li>Second pass (every 3rd): replaces current coin \u2192 D<\/li>\n\n\n\n<li>Third pass (every 4th): replaces current coin \u2192 Q<\/li>\n<\/ul>\n\n\n\n<p>So the <strong>last operation determines the final coin type<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Total Coins:<\/h3>\n\n\n\n<p>We need to count how many coins ended as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pennies: not replaced at all<\/li>\n\n\n\n<li>Nickels: replaced only on the 2nd pass<\/li>\n\n\n\n<li>Dimes: replaced on the 3rd pass but not the 4th<\/li>\n\n\n\n<li>Quarters: replaced on the 4th pass (regardless of previous changes)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Set Definitions:<\/h3>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A = multiples of 2 (nickels): \u230a1001\/2\u230b = 500<\/li>\n\n\n\n<li>B = multiples of 3 (dimes): \u230a1001\/3\u230b = 333<\/li>\n\n\n\n<li>C = multiples of 4 (quarters): \u230a1001\/4\u230b = 250<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s count <strong>how many coins are finally quarters<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Any coin divisible by 4 \u2192 replaced last by a <strong>quarter<\/strong><br>\u2192 So: <strong>250 coins<\/strong> are quarters<\/li>\n<\/ul>\n\n\n\n<p>Now, remove these from the previous groups:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Dimes:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Dime positions = multiples of 3 but <strong>not<\/strong> 4<br>\u2192 Find numbers divisible by 3 and not 4<br>\u2192 Multiples of both 3 and 4 = LCM(3,4) = 12<br>\u2192 \u230a1001\/12\u230b = 83<br>\u2192 Dimes only = 333 &#8211; 83 = <strong>250 coins<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Nickels:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nickel positions = multiples of 2 not divisible by 3 or 4<br>\u2192 Multiples of 6 = LCM(2,3) = \u230a1001\/6\u230b = 166<br>\u2192 Multiples of 4 and 2 = still 250 (already counted in quarters)<br>\u2192 Multiples of 12 = 83<br>\u2192 Nickels only = 500 &#8211; 166 (shared with 3) &#8211; 250 (shared with 4) + 83 (to avoid double subtracting those counted in both 3 and 4)<br>\u2192 Total = 500 &#8211; 166 &#8211; 250 + 83 = <strong>167 coins<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Pennies:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Never touched \u2192 Not divisible by 2, 3, or 4<br>\u2192 Use inclusion-exclusion:<\/li>\n<\/ul>\n\n\n\n<p>Count of numbers from 1 to 1001 not divisible by 2, 3, or 4:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Multiples<\/th><th>Count<\/th><\/tr><\/thead><tbody><tr><td>2<\/td><td>500<\/td><\/tr><tr><td>3<\/td><td>333<\/td><\/tr><tr><td>4<\/td><td>250<\/td><\/tr><tr><td>2 &amp; 3 (6)<\/td><td>166<\/td><\/tr><tr><td>2 &amp; 4 (4)<\/td><td>250<\/td><\/tr><tr><td>3 &amp; 4 (12)<\/td><td>83<\/td><\/tr><tr><td>2,3,4 (LCM=12)<\/td><td>83<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Using inclusion-exclusion:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>Count = 1001 \n        - (Div by 2 + 3 + 4) \n        + (Div by 6 + 8 + 12)\n        - (Div by 12 again since LCM of all three is 12)\n\n= 1001 - (500 + 333 + 250) + (166 + 83 + 83) - 83  \n= 1001 - 1083 + 332 - 83  \n= 1001 - 1083 + 249  \n= 167 pennies<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Tally:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pennies: 167 coins \u2192 167 \u00d7 \\$0.01 = <strong>\\$1.67<\/strong><\/li>\n\n\n\n<li>Nickels: 167 coins \u2192 167 \u00d7 \\$0.05 = <strong>\\$8.35<\/strong><\/li>\n\n\n\n<li>Dimes: 250 coins \u2192 250 \u00d7 \\$0.10 = <strong>\\$25.00<\/strong><\/li>\n\n\n\n<li>Quarters: 250 coins \u2192 250 \u00d7 \\$0.25 = <strong>\\$62.50<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Total Money<\/strong>:<\/h3>\n\n\n\n<p><strong>\\$1.67 + \\$8.35 + \\$25.00 + \\$62.50 = \\$97.52<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer: <strong>\\$97.52<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation Summary (300+ Words):<\/h3>\n\n\n\n<p>This problem involves divisibility and number theory, specifically using inclusion-exclusion principles. Initially, we have 1001 pennies. Replacing every 2nd coin with a nickel means all positions divisible by 2 change to nickels \u2014 that\u2019s 500 coins. But the next step replaces every 3rd coin with a dime, meaning positions divisible by 3 (333 total) are changed again, possibly overwriting nickels. Finally, coins in positions divisible by 4 (250 total) are changed again to quarters, which take final priority.<\/p>\n\n\n\n<p>Because a coin can be replaced multiple times, we need to determine which coins are ultimately left as each type. Any coin that\u2019s divisible by 4 ends as a quarter, regardless of earlier replacements. That\u2019s 250 coins. From the dimes (positions divisible by 3), we subtract those also divisible by 4 (83), leaving 250 coins as final dimes. Nickels are positions divisible by 2 but not by 3 or 4 \u2014 accounting for overlaps using inclusion-exclusion, we get 167 such positions. Finally, the coins never touched are those not divisible by 2, 3, or 4, which again total 167.<\/p>\n\n\n\n<p>Multiplying the count of each type by its value and summing gives the total money:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>167 pennies = \\$1.67<\/li>\n\n\n\n<li>167 nickels = \\$8.35<\/li>\n\n\n\n<li>250 dimes = \\$25.00<\/li>\n\n\n\n<li>250 quarters = \\$62.50<br><strong>Total = \\$97.52<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This problem highlights how overlapping sets require careful accounting using number theory tools, especially when replacements overwrite earlier changes.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>On a table, there are 1001 pennies lined up in a row. I then come along and replace every second penny with a nickel (starting with the second). After this I replace every third coin with a dime (starting with the third). Finally I replace every fourth coin with a quarter (starting with the fourth). [&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-213978","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213978","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=213978"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213978\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213978"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213978"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213978"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}