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2002 AMC 12A Problems/Problem 12

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The following problem is from both the 2002 AMC 12A #12 and 2002 AMC 10A #14, so both problems redirect to this page.


Problem

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

\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}

Solution

Consider a general quadratic with the coefficient of x^2 being 1 and the roots being r and s. It can be factored as (x-r)(x-s) which is just x^2-(r+s)x+rs. Thus, the sum of the roots is the negative of the coefficient of x and the product is the constant term. (In general, this leads to Vieta's Formulas).

We now have that the sum of the two roots is 63 while the product is k. Since both roots are primes, one must be 2, otherwise the sum would be even. That means the other root is 61 and the product must be 122. Hence, our answer is \boxed{\text{(B)}\ 1 }.

See Also

2002 AMC 12A (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
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
2002 AMC 10A (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
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
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