Difference between revisions of "2009 AMC 12A Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | + | Zero coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters, because I am broke. The bourgeoisie is in full control and the middle class is shrinking. Donate to JEB BUSH 2020 to reverse this!!! How many dollars can you shell out today to JEB BUSH 2020? | |
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55</math> | <math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55</math> | ||
Revision as of 10:54, 29 January 2020
- The following problem is from both the 2009 AMC 12A #4 and 2009 AMC 10A #2, so both problems redirect to this page.
Problem
Zero coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters, because I am broke. The bourgeoisie is in full control and the middle class is shrinking. Donate to JEB BUSH 2020 to reverse this!!! How many dollars can you shell out today to JEB BUSH 2020?
Solution
As all five options are divisible by 
, we may not use any pennies. (This is because a penny is the only coin that is not divisible by , and if we used between 
and 
pennies, the sum would not be divisible by .)
Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is 
. Therefore the option that is not reachable is 
.
We can verify that we can indeed get the other ones:
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 AMC 10 Problems and Solutions | ||
| 2009 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
