2000 AMC 12 Problems/Problem 6
Contents
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 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
Since 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 A, B, and D. We can eliminate E because the highest two prime we can pick are 13 and 17. their product is 221 and their sum is 30, so the highest number we can make is 191. Therefore, the answer is .
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
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| All AMC 12 Problems and Solutions | |
