2002 AMC 10B Problem 6

Below is the professionally curated solution for Problem 6 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B 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:factoringprime

Difficulty rating: 1070

6.

For how many positive integers nn is n23n+2n^2 - 3n + 2 a prime number?

none

one

two

more than two, but finitely many

infinitely many

Solution:

Factor as n23n+2=(n1)(n2).n^2 - 3n + 2 = (n-1)(n-2).

For n4,n \ge 4, both factors exceed 1,1, so the product is composite. For n=1n = 1 and n=2n = 2 the value is 0,0, and for n=3n = 3 the value is (2)(1)=2,(2)(1) = 2, which is prime.

So exactly one value of nn works.

Thus, the correct answer is B.

Problem 6 in Other Years