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>.
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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?

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

Solution

As all five options are divisible by $5$, we may not use any pennies. (This is because a penny is the only coin that is not divisible by $5$, and if we used between $1$ and $4$ pennies, the sum would not be divisible by $5$.)

Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is $4\cdot 5 = 20$. Therefore the option that is not reachable is $\boxed{15}$ $\Rightarrow$ $(A)$.

We can verify that we can indeed get the other ones:

  • $25 = 10+5+5+5$
  • $35 = 10+10+10+5$
  • $45 = 25+10+5+5$
  • $55 = 25+10+10+10$

See Also

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
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
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