Difference between revisions of "1989 AHSME Problems/Problem 8"
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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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By expansion of the product of the linear factors and comparison to the original quadratic, | By expansion of the product of the linear factors and comparison to the original quadratic, | ||
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<math>ab = -n</math> | <math>ab = -n</math> | ||
<math>a + b = 1</math>. | <math>a + b = 1</math>. | ||
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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: | ||
<center> | <center> | ||
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<math>a = -1, b = 2</math> | <math>a = -1, b = 2</math> | ||
<math>a = -2, b = 3</math> | <math>a = -2, b = 3</math> | ||
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<math>\vdots</math> | <math>\vdots</math> | ||
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<math>a = -9, b = 10</math> | <math>a = -9, b = 10</math> | ||
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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>. | ||
| + | <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 } </math> | ||
| − | + | == See also == | |
| + | {{AHSME box|year=1989|num-b=7|num-a=9}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 07:45, 22 October 2014
Problem
For how many integers 
between 1 and 100 does 
factor into the product of two linear factors with integer coefficients?
Solution
For to factor into a product of two linear factors, we must have 
, where 
and 
are integers.
By expansion of the product of the linear factors and comparison to the original quadratic,
.
The only way for this to work if is a positive integer is if 
.
Here are the possible pairs:
This gives us 9 integers for , 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
