Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | You have to have 4 pennies so you can have 4 cents, 9 cents, 14, etc. Then you start with 1 dime, 1 nickel, and 1 quarter. You go through 1-9, 10-19, etc. until 90-99, and along the way, you add coins with the most value. You end up with 4 pennies, 1 nickel, 2 dimes, and 3 quarters. <math>4+1+2+3=10</math>, so our answer is <math>\boxed{ | + | You have to have 4 pennies so you can have 4 cents, 9 cents, 14, etc. Then you start with 1 dime, 1 nickel, and 1 quarter. You go through 1-9, 10-19, etc. until 90-99, and along the way, you add coins with the most value. You end up with 4 pennies, 1 nickel, 2 dimes, and 3 quarters. <math>4+1+2+3=10</math>, so our answer is <math>\boxed{\textbf{(B)}\ 10}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=6|num-a=8}} | {{AMC8 box|year=2010|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:56, 21 October 2017
Problem
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
Solution
You have to have 4 pennies so you can have 4 cents, 9 cents, 14, etc. Then you start with 1 dime, 1 nickel, and 1 quarter. You go through 1-9, 10-19, etc. until 90-99, and along the way, you add coins with the most value. You end up with 4 pennies, 1 nickel, 2 dimes, and 3 quarters. 
, so our answer is 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
