Difference between revisions of "2021 JMPSC Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | Brady has an unlimited supply of quarters (<math>0.25), dimes (< | + | Brady has an unlimited supply of quarters (<math>\$0.25</math>), dimes (<math>\$0.10</math>), nickels (<math>\$0.05</math>), and pennies (<math>\$0.01</math>). What is the least number (quantity, not type) of coins Brady can use to pay off <math>\$2.78</math>? |
== Solution == | == Solution == | ||
To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth <math>\$2.75</math>, which is the most quarters we can use to get a value less than or equal to <math>\$2.78</math>. Finally, we can add 3 pennies to get a total of <math>\$2.87</math>, so the answer is <math>11+3=14</math>. | To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth <math>\$2.75</math>, which is the most quarters we can use to get a value less than or equal to <math>\$2.78</math>. Finally, we can add 3 pennies to get a total of <math>\$2.87</math>, so the answer is <math>11+3=14</math>. | ||
Latest revision as of 19:30, 10 July 2021
Problem
Brady has an unlimited supply of quarters (
), dimes (
), nickels (
), and pennies (
). What is the least number (quantity, not type) of coins Brady can use to pay off 
?
Solution
To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth 
, which is the most quarters we can use to get a value less than or equal to . Finally, we can add 3 pennies to get a total of 
, so the answer is 
.
