Difference between revisions of "1989 AHSME Problems/Problem 8"

Revision as of 12:20, 21 May 2014

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.

@@ Line 1: / Line 1: @@
 For how many integers <math>n</math> between 1 and 100 does <math>x^2+x-n</math> factor into the product of two linear factors with integer coefficients?
+For <math>x^2+x-n</math> to factor into a product of two linear factors, we must have <math>x^2+x-n = (x + a)(x + b)</math>, where <math>a</math> and <math>b</math> are integers.
+By expansion of the product of the linear factors and comparison to the original quadratic,
+<math>ab = -n</math>
+<math>a + b = 1</math>.
+The only way for this to work if n is a positive integer is if <math>a = -b +1</math>.
+Here are the possible pairs:
+<center>
+<math>a = -1, b = 2</math>
+<math>a = -2, b = 3</math>
+<math>\vdots</math>
+<math>a = -9, b = 10</math>
+</center>
+This gives us 9 integers for <math>n</math>, <math>\boxed{\text{D}}</math>.
 <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  </math>
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