{"id":37571,"date":"2025-06-26T02:42:51","date_gmt":"2025-06-26T02:42:51","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37571"},"modified":"2025-06-26T02:42:52","modified_gmt":"2025-06-26T02:42:52","slug":"how-many-ways-can-change-be-made-for-one-dollar-using-each-of-the-following-coins","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-ways-can-change-be-made-for-one-dollar-using-each-of-the-following-coins\/","title":{"rendered":"How many ways can change be made for one dollar, using each of the following coins"},"content":{"rendered":"\n
How many ways can change be made for one dollar, using each of the following coins? (a) dimes and quarters (b) nickels, dimes, and quarters (c) pennies, nickels, dimes, and quarters.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let\u2019s solve the problem of finding how many ways to make change for one dollar (100 cents) using different sets of coins:<\/p>\n\n\n\n
\n\n\n\n
(a) Dimes and Quarters<\/strong><\/h3>\n\n\n\n
Let ddd be the number of dimes (10 cents) and qqq be the number of quarters (25 cents). We want:10d+25q=10010d + 25q = 10010d+25q=100<\/p>\n\n\n\n
Divide both sides by 5:2d+5q=202d + 5q = 202d+5q=20<\/p>\n\n\n\n
Now solve for integer solutions. Try values of qqq from 0 to 4 (since 5 quarters = 125 > 100):<\/p>\n\n\n\n
q (quarters)<\/th>
Equation: 2d = 20 – 5q<\/th>
d (dimes)<\/th><\/tr><\/thead>
0<\/td>
2d = 20<\/td>
d = 10<\/td><\/tr>
1<\/td>
2d = 15<\/td>
Not integer<\/td><\/tr>
2<\/td>
2d = 10<\/td>
d = 5<\/td><\/tr>
3<\/td>
2d = 5<\/td>
Not integer<\/td><\/tr>
4<\/td>
2d = 0<\/td>
d = 0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
(b) Nickels, Dimes, and Quarters<\/strong><\/h3>\n\n\n\n
Let nnn, ddd, and qqq represent nickels (5\u00a2), dimes (10\u00a2), and quarters (25\u00a2) respectively:5n+10d+25q=1005n + 10d + 25q = 1005n+10d+25q=100<\/p>\n\n\n\n
Divide by 5:n+2d+5q=20n + 2d + 5q = 20n+2d+5q=20<\/p>\n\n\n\n
Now iterate over values of qqq from 0 to 4 (same reasoning), for each find valid (d,n)(d, n)(d,n) such that:n+2d=20\u22125qn + 2d = 20 – 5qn+2d=20\u22125q<\/p>\n\n\n\n
(c) Pennies, Nickels, Dimes, and Quarters<\/strong><\/h3>\n\n\n\n
Let p,n,d,qp, n, d, qp,n,d,q be the number of pennies (1\u00a2), nickels (5\u00a2), dimes (10\u00a2), and quarters (25\u00a2). We solve:p+5n+10d+25q=100p + 5n + 10d + 25q = 100p+5n+10d+25q=100<\/p>\n\n\n\n
Let\u2019s fix values of qqq from 0 to 4, then for each, loop through values of ddd and nnn, and compute the value of p=100\u2212(25q+10d+5n)p = 100 – (25q + 10d + 5n)p=100\u2212(25q+10d+5n). We count all cases where p\u22650p \\ge 0p\u22650 and integer.<\/p>\n\n\n\n
Using a computer or generating function or direct counting, this results in:<\/p>\n\n\n\n
\u2705 Total combinations: 242<\/strong><\/p>\n\n\n\n\n\n\n\n
Summary<\/strong><\/h3>\n\n\n\n
\n
(a) Dimes and Quarters: 3 ways<\/strong><\/li>\n\n\n\n
(b) Nickels, Dimes, and Quarters: 29 ways<\/strong><\/li>\n\n\n\n
(c) Pennies, Nickels, Dimes, and Quarters: 242 ways<\/strong><\/li>\n<\/ul>\n\n\n\n
Each step involves counting integer solutions to linear Diophantine equations under coin constraints. Including more coin types increases the number of combinations significantly.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
How many ways can change be made for one dollar, using each of the following coins? (a) dimes and quarters (b) nickels, dimes, and quarters (c) pennies, nickels, dimes, and quarters. The Correct Answer and Explanation is: Let\u2019s solve the problem of finding how many ways to make change for one dollar (100 cents) using […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-37571","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37571","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=37571"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37571\/revisions"}],"predecessor-version":[{"id":37573,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37571\/revisions\/37573"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37571"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37571"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"