Difference between revisions of "2021 JMPSC Sprint Problems/Problem 2"
| Line 11: | Line 11: | ||
== 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> | ||
| + | |||
| + | - 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. 
