Question

正整数 $A$ 、 $B$ 、 $A - B$ 和 $A + B$ 都是质数。这四个质数之和

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

是偶数

even

能被 $3$ 整除

divisible by $3$

能被 $5$ 整除

divisible by $5$

能被 $7$ 整除

divisible by $7$

是质数

prime

答案：E

知识点：质数奇偶性

难度评级：1480

解答：

数 $A - B$ 与 $A + B$ 相差 $2B,$ 所以奇偶性相同。因为它们是质数，两者都必须是奇数，这迫使 $A$ 与 $B$ 奇偶性相反。

因为 $2$ 是唯一的偶质数，所以要么 $A=2$ ，要么 $B=2.$ 第一种情况不可能，因为正质数 $B$ 会使 $A-B<0.$ 因此 $B=2.$

现在 $A - 2,$ $A,$ 和 $A + 2$ 是三个质数。任意三个相差 $2$ 的整数中必有一个能被 $3$ 整除，所以这个数本身必须是 $3.$ 唯一的正数情形是 $3,5,7.$

这四个质数是 $2, 3, 5, 7,$ 它们的和为 $17,$ 也是质数。

所以正确答案是 E。

The numbers $A - B$ and $A + B$ differ by $2B,$ so they have the same parity. Being prime, they must both be odd, which forces $A$ and $B$ to have opposite parity.

Since $2$ is the only even prime, either $A=2$ or $B=2.$ The first case is impossible because the positive prime $B$ would make $A-B<0.$ Hence $B=2.$

Now $A - 2,$ $A,$ and $A + 2$ are three primes. One of any three integers spaced $2$ apart is divisible by $3,$ so that member must itself be $3.$ The only positive possibility is $3,5,7.$

The four primes are $2, 3, 5, 7,$ and their sum is $17,$ which is prime.

Thus, the correct answer is E.

2002 AMC 10B 第 15 题

答案：E

难度评级：1480

解答：

其他年份的第 15 题