Difference between revisions of "2009 AMC 12A Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | + | Note that if we use any pennies whatsoever (one, two, three, or four of them), then the total amount of money would be nonzero mod <math>5</math>; as all five options are divisible by <math>5</math>, we conclude that we may not use any pennies. | |
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>. | 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>. | ||
| Line 17: | Line 17: | ||
* <math>45 = 25+10+5+5</math> | * <math>45 = 25+10+5+5</math> | ||
* <math>55 = 25+10+10+10</math> | * <math>55 = 25+10+10+10</math> | ||
| + | |||
| + | (Minor clarity edit by Technodoggo) | ||
== 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}} | {{MAA Notice}} | ||
Latest revision as of 23:07, 31 August 2024
- 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
Note that if we use any pennies whatsoever (one, two, three, or four of them), then the total amount of money would be nonzero mod 
; as all five options are divisible by , we conclude that we may not use any pennies.
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:
(Minor clarity edit by Technodoggo)
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. 
