2000 AMC 12 Problems/Problem 6

The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.

Problem

Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

Solution

Solution 1

Let the primes be and .

The problem asks us for possible values of where

Using Simon's Favorite Factoring Trick:

Possible values of and are:

The possible values for (formed by multipling two distinct values for and ) are:

So the possible values of are:

The only answer choice on this list is

Note: once we apply the factoring trick we see that, since and are even, should be a multiple of .

These means that only and are possible.

We can't have with and below . Indeed, would have to be or .

But could be or Of these, three have and prime, but only the last has them both small enough. Therefore the answer is .

Solution 2

All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is . Thus, we can eliminate E. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is . Therefore, A cannot be an answer. So, the answer must be .

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.