2000 AMC 10 Problems/Problem 11

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.

odd*odd=odd.

odd-odd-odd=odd.

Thus, we can discard the even choices.

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

$(a-1)(b-1)-1=21,119,231$ .

Both of these factors are even, so the number +1 must be a multiple of $4$ .

$120$ is the only possiblity.

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

C

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

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