2002 AMC 10B Problem 6

Try a few values first to see what happens:

Every value is even. Here is why.

1. Factor: $n^2 - 3n + 2 = (n - 1)(n - 2)$.
2. The numbers $n - 2$ and $n - 1$ are next to each other on the number line, so one of them is even. That makes their product even every time.
3. The only even prime is $2$. So the expression can be prime only when it equals $2$.
4. Two neighboring whole numbers that multiply to $2$ are $1$ and $2$ (so $n = 3$), or $-2$ and $-1$ (so $n = 0$). But $0$ is not a positive integer.
5. That leaves only $n = 3$, and $9 - 9 + 2 = 2$ is indeed prime.

The answer is $\boxed{\textbf{(B)}\ \text{one}}$.

If a quadratic factors into two whole-number pieces, it can only be prime when one piece is $1$ or $-1$. Here there was an even quicker check: two neighbors always include an even number, so the product is always even and $2$ is the only prime it can hit. Factor first, then ask what the pieces are allowed to be.

Skip the even-number idea. Once $n \ge 4$, the factor $n - 2$ is at least $2$ and $n - 1$ is at least $3$, so their product has two factors bigger than $1$ and cannot be prime. That leaves $n = 1, 2, 3$, which give $0, 0, 2$. Exactly one prime.

Factor the quadratic: $n^2 - 3n + 2 = (n-1)(n-2)$.

The factors $n - 2$ and $n - 1$ are consecutive integers, so one of them is even and the product is always even. The only even prime is $2$, so we need
$$(n-1)(n-2) = 2.$$
Two consecutive integers with product $2$ are $1$ and $2$ (giving $n = 3$) or $-2$ and $-1$ (giving $n = 0$, which is not positive). Only $n = 3$ survives, and indeed $9 - 9 + 2 = 2$ is prime.

The answer is $\boxed{\textbf{(B)}\ \text{one}}$.