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

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==Problem==
 
==Problem==
 
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 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?
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<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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==Solution==
 
==Solution==
 
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.
 
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.
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The only way for this to work if <math>n</math> is a positive integer is if <math>a = -b +1</math>.
 
The only way for this to work if <math>n</math> is a positive integer is if <math>a = -b +1</math>.
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Here are the possible pairs:
 
Here are the possible pairs:
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This gives us 9 integers for <math>n</math>, <math>\boxed{\text{D}}</math>.
 
This gives us 9 integers for <math>n</math>, <math>\boxed{\text{D}}</math>.
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<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  </math>
 
  
 
== See also ==
 
== See also ==

Revision as of 06:46, 22 October 2014

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?

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$

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 $n$, $\boxed{\text{D}}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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