Difference between revisions of "2009 AMC 12A Problems/Problem 4"
(New 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 <em>not</em> be the total value of...) |
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| + | {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #4]] and [[2009 AMC 10A Problems|2009 AMC 10A #2]]}} | ||
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== Problem == | == 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 <em>not</em> be the total value of the four coins, in cents? | Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could <em>not</em> be the total value of the four coins, in cents? | ||
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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: | ||
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== See Also == | == See Also == | ||
| + | {{AMC10 box|year=2009|ab=A|num-b=1|num-a=3}} | ||
{{AMC12 box|year=2009|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2009|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:40, 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
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 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. 
