# Difference between revisions of "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