AMC12 2002 B

AMC12 2002 B · Q3

AMC12 2002 B · Q3. It mainly tests Quadratic equations, Primes & prime factorization.

For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?

有多少个正整数 $n$ 使得 $n^2 - 3n + 2$ 是质数？

(A) none 没有

(B) one 一个

(D) more than two, but finitely many 超过两个，但有限个

(E) infinitely many 无限多个

Answer

Correct choice: (B)

正确答案：（B）

Solution

Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$. Either $n-1$ or $n-2$ is odd, and the other is even. Their product must yield an even number. The only prime that is even is $2$, which is when $n$ is $3$ or $0$. Since $0$ is not a positive number, the answer is $\boxed{\mathrm{(B)}\ \text{one}}$.

因式分解得 $n^2 - 3n + 2 = (n-2)(n-1)$。$n-1$ 与 $n-2$ 中一个为奇数、另一个为偶数，它们的乘积必为偶数。唯一的偶质数是 $2$，这发生在 $n$ 为 $3$ 或 $0$ 时。由于 $0$ 不是正数，答案是 $\boxed{\mathrm{(B)}\ \text{one}}$。

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