AMC12 2009 B

AMC12 2009 B · Q19

AMC12 2009 B · Q19. It mainly tests Quadratic equations, Primes & prime factorization.

For each positive integer $n$, let $f(n) = n^4 - 360n^2 + 400$. What is the sum of all values of $f(n)$ that are prime numbers?

对每个正整数 $n$，令 $f(n) = n^4 - 360n^2 + 400$。所有使得 $f(n)$ 为素数的取值中，$f(n)$ 的总和是多少？

(A) 794 794

(B) 796 796

(D) 800 800

(E) 802 802

Answer

Correct choice: (E)

正确答案：（E）

Solution

To find the answer it was enough to play around with $f$. One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$, and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$.

为了得到答案，只需对 $f$ 做一些尝试。容易发现 $f(1)=41$ 是素数；随后当 $n$ 在 $2$ 到 $18$ 之间时 $f$ 为负；接着 $f(19)=761$ 又是素数。并且由于 $41 + 761 = 802$ 已经是选项中最大的一个，因此答案必为 $\boxed{802}$。

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