2002 AMC 12A Problems/Problem 12

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

Solution 2

Let the quadratic be factored as $(x-r)(x-s)$ . Then $-r-s=-63$ and $rs=k$ . We know that both $r$ and $s$ are prime, and all primes but $2$ are odd. The sum of two odd numbers is also always even, but in this case the sum of two primes is odd, which means that one of the primes is $2$ . This leaves only one possible value for the other root and one possible value for their product $k$ .

Solution 3

By Vieta's you have $$ (Error compiling LaTeX. Unknown error_msg)a+b=63 $<cmath>ab=k</cmath> If$ a $and$ b$ are both primes and sum to 63 there must be only 1 way because we know odd + even = odd, and 2 is the only even prime => 2 + 61 = 63, both are prime, so 1 way.

Video Solution

https://youtu.be/5QdPQ3__a7I?t=130

~ pi_is_3.14

2002 AMC 12A (Problems • Answer Key • Resources)
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
All AMC 12 Problems and Solutions

2002 AMC 10A (Problems • Answer Key • Resources)
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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