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 10:09, 12 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

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 $11$. Since you can't use three cents on anything other than pennies, the remaining coins are $3$ pennies. Therefore $11+3=14$

- kante314 -

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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