Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99 </math> | <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99 </math> | ||
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| + | ==Solution== | ||
| + | If he has <math>3</math> quarters, his other coins are able to pay amount amount less than or equal to <math>25</math> cents, because it can just be added on to the value of his quarters. If he has <math>2</math> dimes and <math>1</math> nickel, he can pay any amount that is a multiple of <math>5</math>. Lastly, Freddie need <math>4</math> pennies for a total of <math>3+2+1+4=\boxed{\textbf{(B)}\ 10}</math> 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}} | ||
Revision as of 18:01, 25 December 2012
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
If he has 
quarters, his other coins are able to pay amount amount less than or equal to 
cents, because it can just be added on to the value of his quarters. If he has 
dimes and 
nickel, he can pay any amount that is a multiple of 
. Lastly, Freddie need 
pennies for a total of 
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 | ||
