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2002 AMC 10A Problem 14

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2002 AMC 10A Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Primes · difficulty 3 of 5 · about 3 min · position P11-15

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

A)
$0$
B)
$1$
C)
$2$
D)
$4$
E)
$more than 4$

Answer

Key insight

The roots add to the odd number 63, so one prime root must be the only even prime, 2, forcing the other to be 61.

Solution

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 $\boxed{\textbf{(B)}\ 1}$ .

Why this works

Vieta turns a statement about roots into a statement about a sum and a product. Parity then does the rest: any time two primes must add to an odd number, one of them is $2$ . This "odd sum means a 2 is present" step recurs constantly in AMC number theory.

The trap

Searching for prime pairs summing to 63 without the parity shortcut, or counting (2, 61) and (61, 2) as two different values of k.

Common mistakes

Searching for prime pairs summing to 63 without the parity shortcut, or counting (2, 61) and (61, 2) as two different values of k.
Forgetting to verify that $61$ is prime before concluding.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Use an invariant, parity, or coloring argument