Art of Problem Solving

2021 JMPSC Sprint Problems/Problem 2: Difference between revisions

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==Solution==
==Solution==
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It is generally best to use the smallest number of coins with the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit <math>11</math> quarters, or <math>\$2.75</math> into <math>\$2.78</math>. That only leaves <math>3</math> cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of <math>\$2.78</math>.
 
The least number of coins Brady can use to pay off <math>\$2.78</math> will be <math>14</math> coins.
 
-OofPirate

Revision as of 23:44, 10 July 2021

Problem

Brady has an unlimited supply of quarters (\$0.25), dimes (\$0.10), nickels (\$0.05), and pennies (\$0.01). What is the least number (quantity, not type) of coins Brady can use to pay off \$$2.78$?

Solution

It is generally best to use the smallest number of coins with the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit $11$ quarters, or $\$2.75$ into $\$2.78$. That only leaves $3$ cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of $\$2.78$.

The least number of coins Brady can use to pay off $\$2.78$ will be $14$ coins.

-OofPirate