2002 AMC 10A Problem 14 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10A Problem 14

Below is the professionally curated solution for Problem 14 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:Vieta’s Formulas prime parity

Difficulty rating: 1310

14.

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

$0$

$1$

$2$

$4$

more than four

Solution:

If the roots are primes $p$ and $q,$ then $p+q=63$ and $pq=k.$ Because $63$ is odd, one prime must be $2,$ making the other $61,$ which is prime.

So $k=2\cdot 61=122$ is the only possible value.

Thus, the correct answer is B.

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