Difference between revisions of "2009 AMC 12A Problems/Problem 4"
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As all five options are divisible by <math>5</math>, we may not use any pennies. (This is because a penny is the only coin that is not divisible by <math>5</math>, and if we used between <math>1</math> and <math>4</math> pennies, the sum would not be divisible by <math>5</math>.) | As all five options are divisible by <math>5</math>, we may not use any pennies. (This is because a penny is the only coin that is not divisible by <math>5</math>, and if we used between <math>1</math> and <math>4</math> pennies, the sum would not be divisible by <math>5</math>.) | ||
− | Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is <math>4\cdot 5 = 20</math>. Therefore the option that is not reachable is <math>\boxed{15}</math>. | + | Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is <math>4\cdot 5 = 20</math>. Therefore the option that is not reachable is <math>\boxed{15}</math> <math>\Rightarrow</math> <math>(A)</math>. |
We can verify that we can indeed get the other ones: | We can verify that we can indeed get the other ones: |
Revision as of 15:31, 1 January 2012
- The following problem is from both the 2009 AMC 12A #4 and 2009 AMC 10A #2, so both problems redirect to this page.
Problem
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
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 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 |
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 |