Difference between revisions of "2010 AMC 8 Problems/Problem 7"
(→Solution) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | To have less then a dollar, the most it could be is 99 cents. You could use 3 of the quarters, then 2 dimes, and 4 pennies to make 99 cents with least coins. | + | To have less then a dollar, the most it could be is 99 cents. You could use 3 of the quarters, then 2 dimes, and 4 pennies to make 99 cents with least coins. So the total of coins is 6 coins. |
==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 19:24, 20 September 2015
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
To have less then a dollar, the most it could be is 99 cents. You could use 3 of the quarters, then 2 dimes, and 4 pennies to make 99 cents with least coins. So the total of coins is 6 coins.
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. 
