AMC12 2000 A

AMC12 2000 A · Q6

AMC12 2000 A · Q6. 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 numbers could be obtained?

在 $4$ 到 $18$ 之间选取两个不同的质数。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

Any two 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. 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. So, the answer must be $\boxed{\textbf{(C) }119}$.

在 4 到 18 之间任取两个质数，它们的积为奇数，和为偶数。奇数减去偶数仍为奇数，因此可排除 A、B 和 D。由于能选取的最大两个质数是 13 和 17，所能得到的最大值为 $(13)(17)-(13+17) = 221 - 30 = 191$，因此可排除 E。故答案为 $\boxed{\textbf{(C) }119}$。

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