# Difference between revisions of "2021 JMPSC Sprint Problems/Problem 2"

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== Solution 2 == | == Solution 2 == | ||

You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is <math>11</math>. Since you can't use three cents on anything other than pennies, the remaining coins are <math>3</math> pennies. Therefore <math>11+3=14</math> | You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is <math>11</math>. Since you can't use three cents on anything other than pennies, the remaining coins are <math>3</math> pennies. Therefore <math>11+3=14</math> | ||

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+ | - kante314 - | ||

==See also== | ==See also== |

## Latest revision as of 09:09, 12 July 2021

## Contents

## 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 $?

## 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 quarters, or into . That only leaves cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of .

The least number of coins Brady can use to pay off will be coins.

-OofPirate

## Solution 2

You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is . Since you can't use three cents on anything other than pennies, the remaining coins are pennies. Therefore

- kante314 -

## See also

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.