1989 AHSME Problems/Problem 8

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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?

Solution

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. AMC logo.png