2000 AMC 12 Problems/Problem 6

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

Let the primes be $p$ and $q$ .

The problem asks us for possible values of $K$ where $K=pq-p-q$

Using Simon's Favorite Factoring Trick:

$K+1=pq-p-q+1$

$K+1=(p-1)(q-1)$

Possible values of $(p-1)$ and $(q-1)$ are:

$4,6,10,12,16$

The possible values for $K+1$ (formed by multipling two distinct values for $(p-1)$ and $(q-1)$ ) are:

$24,40,48,60,64,72,96,120,160,192$

So the possible values of $K$ are:

$23,39,47,59,63,71,95,119,159,191$

The only answer choice on this list is $119 \Rightarrow C$

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2000 AMC 12 Problems/Problem 6

Problem

Solution

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