AMC12每日一题(2000年真题#06)

2000 AMC 12竞赛试题/第6题

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?

           (A)   21       (B)   60         (C)    119           (D)  180          (E)  231

Solution 1

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(13)(17)-(13+17)=221-30=191. 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(5)(7)-(5+7)=23..Therefore, A cannot be an answer. So, the answer must be (C).

Solution 2

Let the two primes be P and q. We wish to obtain the value of pq-(p+q), or pq-p-q. Using Simon's Favorite Factoring Trick, we can rewrite this expression as(1-p)(1-q)-1 or(p-1)(q-1)-1 . Noticing that(13-1)(11-1)-1=120-1=119 , we see that the answer is (C).