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

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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>
 
<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==
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==Solution 1==
 
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.
  

Latest revision as of 13:24, 27 June 2021

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 1

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

Solution 2

For $x^2+x-n$ to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals $\sqrt{1^2-4(1)(-n)}$ or $\sqrt{4n+1}$.

Since $n$ must be a positive integer, $4n+1$ must be odd because if it were even, $4n$ would have to be both odd and divisible by 4, which is a contradiction. Therefore, $n$ must be even.

The maximum value of $4n+1$ is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401. There are 9: $9,25,49,81,121,169,225,289,$and $361$. Therefore, the answer is $\boxed{D}$.

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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