1989 AHSME Problems/Problem 8

Problem

For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?

For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$ , where $a$ and $b$ are integers.

By expansion of the product of the linear factors and comparison to the original quadratic,

$ab = -n$

$a + b = 1$ .

The only way for this to work if $n$ is a positive integer is if $a = -b +1$ .

Here are the possible pairs:

$a = -1, b = 2$

$a = -2, b = 3$

$\vdots$

$a = -9, b = 10$

This gives us 9 integers for $n$ , $\boxed{\text{D}}$ .

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.