# 2002 AMC 12B Problems/Problem 3

The following problem is from both the 2002 AMC 12B #3 and 2002 AMC 10B #6, so both problems redirect to this page.

## Problem

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

$\mathrm{(A)}\ \text{none} \qquad\mathrm{(B)}\ \text{one} \qquad\mathrm{(C)}\ \text{two} \qquad\mathrm{(D)}\ \text{more\ than\ two,\ but\ finitely\ many} \qquad\mathrm{(E)}\ \text{infinitely\ many}$

## Solution 1

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 $\boxed{\mathrm{(B)}\ \text{one}}$.

## Solution 2

Considering parity, we see that $n^2 - 3n + 2$ is always even. The only even prime is $2$, and so $n^2-3n=0$ whence $n=3\Rightarrow\boxed{\mathrm{(B)}\ \text{one}}$.

 2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2002 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions