2009 AMC 12B Problem 19 — Solution | LIVE by Po-Shen Loh

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2009 AMC 12B Problem 19

Below is the professionally curated solution for Problem 19 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:difference of squares factoring prime

Difficulty rating: 2000

19.

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?

$794$

$796$

$798$

$800$

$802$

Solution:

Write $f(n) = n^4 + 40n^2 + 400 - 400n^2 = (n^2 + 20)^2 - (20n)^2 = (n^2 + 20n + 20)(n^2 - 20n + 20).$

For $f(n)$ to be prime the smaller factor must be $1$ : solving $n^2 - 20n + 20 = 1$ gives $(n - 1)(n - 19) = 0,$ so $n = 1$ or $n = 19.$

Then $f(1) = 41$ and $f(19) = 761$ are both prime, summing to $802.$

Thus, the correct answer is E.

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