Difference between revisions of "2009 AMC 12A Problems/Problem 4"

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<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>
 
== 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>.
 
 
We can verify that we can indeed get the other ones:
 
* <math>25 = 10+5+5+5</math>
 
* <math>35 = 10+10+10+5</math>
 
* <math>45 = 25+10+5+5</math>
 
* <math>55 = 25+10+10+10</math>
 
 
(Minor clarity edit by Technodoggo)
 
  
 
== See Also ==
 
== See Also ==

Revision as of 20:49, 23 May 2025

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?

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55$

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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