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 $119 \Rightarrow C$

Note: once we apply the factoring trick we see that, since $p-1$ and $q-1$ are even, $K+1$ should be a multiple of $4$ .

These means that only $119 \Rightarrow C$ and $231 \Rightarrow E$ are possible.

We can't have $(p-1) \cdot (q-1)=232=2^3\cdot 29$ with $p$ and $q$ below $18$ . Indeed, $(p-1) \cdot (q-1)$ would have to be $2 \cdot 116$ or $4 \cdot 58$ .

But $(p-1) \cdot (q-1)=120=2^3\cdot 3 \cdot 5$ could be $2 \cdot 60,4 \cdot 30,6 \cdot 20$ or $10 \cdot 12.$ Of these, three have $p$ and $q$ prime, but only the last has them both small enough. Therefore the answer is $119 C$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 12 Problems/Problem 6

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