AMC12 2002 B

AMC12 2002 B · Q11

AMC12 2002 B · Q11. It mainly tests Primes & prime factorization, Parity (odd/even).

The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is

正整数 $A, B, A-B,$ 和 $A+B$ 都是素数。这四个素数的和是

(A) even 偶数

(B) divisible by 3 能被3整除

(D) divisible by 7 能被7整除

(E) prime 素数

Answer

Correct choice: (E)

正确答案：（E）

Solution

Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $A-B$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$, it follows that $A$ (as a prime greater than $2$) is odd. Thus $B = 2$, and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$, from which it follows that $A-2 = 3$ and $A = 5$. The sum of these numbers is thus $17$, a prime, so the answer is $\boxed{\mathrm{(E)}\ \text{prime}}$.

由于 $A-B$ 和 $A+B$ 必须同奇偶，而偶素数只有一个，因此 $A-B$ 和 $A+B$ 都是奇数。于是 $A, B$ 中一个为奇数、另一个为偶数。又因为 $A+B > A > A-B > 2$，所以 $A$（作为大于 $2$ 的素数）为奇数。因此 $B=2$，并且 $A-2, A, A+2$ 是连续的奇素数。$A-2, A, A+2$ 中至少有一个能被 $3$ 整除，从而得到 $A-2=3$ 且 $A=5$。这四个数之和为 $17$，是素数，所以答案是 $\boxed{\mathrm{(E)}\ \text{prime}}$。

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