AMC10 2002 B

AMC10 2002 B · Q6

AMC10 2002 B · Q6. It mainly tests Quadratic equations, Primes & prime factorization.

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

有且仅有几个正整数 $n$ 使得 $n^2 -3n + 2$ 是质数？

(A) none 无

(B) one 一个

(D) more than two, but finitely many 有限多个但超过两个

(E) infinitely many 无限多个

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) If $n \ge 4$, then $n^2 - 3n + 2 = (n - 1)(n - 2)$ is the product of two integers greater than 1, and thus is not prime. For $n = 1$, $2$, and $3$ we have, respectively, $(1 - 1)(1 - 2) = 0$, $(2 - 1)(2 - 2) = 0$, and $(3 - 1)(3 - 2) = 2$. Therefore, $n^2 - 3n + 2$ is prime only when $n = 3$.

(B) 若 $n \ge 4$，则 $n^2 - 3n + 2 = (n - 1)(n - 2)$ 是两个大于 1 的整数的乘积，因此不是素数。对 $n = 1、2、3$ 分别有：$(1 - 1)(1 - 2) = 0$，$(2 - 1)(2 - 2) = 0$，以及 $(3 - 1)(3 - 2) = 2$。因此，$n^2 - 3n + 2$ 只有在 $n = 3$ 时才是素数。

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