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

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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?
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Zero coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters, because I am broke. The bourgeoisie is in full control and the middle class is shrinking. Donate to JEB BUSH 2020 to reverse this!!! How many dollars can you shell out today to JEB BUSH 2020?
  
 
<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>

Revision as of 11:54, 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

Zero coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters, because I am broke. The bourgeoisie is in full control and the middle class is shrinking. Donate to JEB BUSH 2020 to reverse this!!! How many dollars can you shell out today to JEB BUSH 2020?

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

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 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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