AMC10 2000 A

AMC10 2000 A · Q11

AMC10 2000 A · Q11. It mainly tests Primes & prime factorization, Parity (odd/even).

Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following number could be obtained?

从4和18之间选择两个不同的质数。当它们的和从它们的积中减去时，下列哪个数可能得到？

(A) 21 21

(B) 60 60

(D) 180 180

(E) 231 231

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): There are five prime numbers between 4 and 18: 5, 7, 11, 13, and 17. Hence the product of any two of these is odd and the sum is even. Because $xy-(x+y)=(x-1)(y-1)-1$ increases as either $x$ or $y$ increases (since both $x$ and $y$ are bigger than 1), the answer must be an odd number that is no smaller than $23=5\cdot 7-(5+7)$ and no larger than $191=13\cdot 17-(13+17)$. The only possibility among the options is 119, and indeed $119=11\cdot 13-(11+13)$.

答案（C）：在 4 和 18 之间有五个质数：5、7、11、13 和 17。因此，其中任意两个的乘积为奇数，而它们的和为偶数。因为 $xy-(x+y)=(x-1)(y-1)-1$ 会随着 $x$ 或 $y$ 的增大而增大（由于 $x$ 和 $y$ 都大于 1），所以答案必须是一个不小于 $23=5\cdot 7-(5+7)$ 且不大于 $191=13\cdot 17-(13+17)$ 的奇数。选项中唯一可能的是 119，且确实有 $119=11\cdot 13-(11+13)$。

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