2000 AMC 10 Problems/Problem 11

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Problem

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$\mathrm{(A)}\ 21 \qquad\mathrm{(B)}\ 60 \qquad\mathrm{(C)}\ 119 \qquad\mathrm{(D)}\ 180 \qquad\mathrm{(E)}\ 231$

Solution

Two prime numbers between $4$ and $18$ are both odd.

$(\text{odd}\cdot\text{odd})-(\text{odd}+\text{odd})=\text{odd}-\text{even}=\text{odd}$

Thus, we can discard the even choices.

$ab-(a+b)=ab-a-b=(a-1)(b-1)-1$

Both $a-1$ and $b-1$ are even, so one more than $ab-a-b$ is a multiple of four.

$119$ is the only possible choice.

$11,13$ satisfy this, $11\cdot 13-11-13=143-24=119$ .

$\boxed{\text{C}}$

2000 AMC 10 (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
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2000 AMC 10 Problems/Problem 11

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