1987 AHSME Problems/Problem 23

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Problem

If $p$ is a prime and both roots of $x^2+px-444p=0$ are integers, then

$\textbf{(A)}\ 1<p\le 11 \qquad \textbf{(B)}\ 11<p \le 21 \qquad \textbf{(C)}\ 21< p \le 31 \\ \qquad \textbf{(D)}\ 31< p\le 41\qquad \textbf{(E)}\ 41< p\le 51$

Solution

For integer roots, we need the discriminant, which is $p^2 - 4 \cdot 1 \cdot (-444p) = p^2 + 1776p = p(p+1776)$, to be a perfect square. Now, this means that $p$ must divide $p + 1776$, as if it did not, there would be a lone prime factor of $p$, and so this expression could not possibly be a perfect square. Thus $p$ divides $p + 1776$, which implies $p$ divides $1776 = 2^{4} \cdot 3 \cdot 37$, so we must have $p = 2$, $3$, or $37$. It is easy to verify that neither $p = 2$ nor $p = 3$ make $p(p+1776)$ a perfect square, but $p = 37$ does, so the answer is $31 < p \le 41$, which is $\boxed{\text{D}}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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