2002 AMC 10A Problem 14

First we need to connect the roots to the numbers $63$ and $k$ in the equation.

1. How roots relate to the equation. Look at an example: $(x - 2)(x - 5) = x^2 - 7x + 10$ has roots $2$ and $5$. The middle number $7$ is $2 + 5$ and the last number $10$ is $2 \times 5$. In the same way, if the roots are $p$ and $q$, then
$$(x - p)(x - q) = x^2 - (p + q)x + pq,$$
so here $p + q = 63$ and $pq = k$.

1. Use odd and even. Odd plus odd is even. But $63$ is odd, so $p$ and $q$ cannot both be odd; one of them is even.

1. The only even prime is $2$. So one root is $2$, and the other is $63 - 2 = 61$.

4. Check. $61$ is prime (it is not divisible by $2$, $3$, $5$ or $7$, and $8^2 = 64$ is already bigger than $61$). So the roots $2$ and $61$ work, and
$$k = 2 \times 61 = 122.$$

1. There was no other choice at any step, so $k = 122$ is the only possible value.

The answer is $\boxed{\textbf{(B)}\ 1}$.

For $x^2 - sx + t = 0$, the roots add to $s$ and multiply to $t$; this turns a question about roots into a question about two numbers. Then odd and even does the rest: whenever two primes add up to an odd number, one of them must be $2$. That shortcut comes up again and again in contest problems.

Let the roots be primes $p$ and $q$. By Vieta, $p + q = 63$ and $pq = k$.

Two odd numbers add to an even number, so a sum of $63$ forces one root to be even. The only even prime is $2$, so $p = 2$ and $q = 61$, which is prime. Hence $k = 2 \cdot 61 = 122$ is the only possibility.

The answer is $\boxed{\textbf{(B)}\ 1}$.