2000 AMC 12 Problem 6

Question

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?

21

60

119

180

231

The primes are

5, 7, 11, 13,

and

17

; note that

xy - (x + y)

= (x - 1)(y - 1) - 1

Both primes are odd, so

(x-1)(y-1)-1 \equiv 3 \pmod 4;

then test the remaining candidates

The primes between

4

and

18

are

5, 7, 11, 13,

and

17.

For two such primes,

xy - (x + y)

= (x - 1)(y - 1) - 1

is a product of two even numbers minus

1,

hence it is

3 \pmod 4.

This leaves

119

and

231.

The latter would require

(x-1)(y-1)=232,

but no two distinct numbers in

\{4,6,10,12,16\}

have product

232.

Indeed,

\begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119. \end{aligned}

Thus, the correct answer is C.

Problem 6 in other years