Difference between revisions of "1989 AHSME Problems/Problem 8"
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+ | ==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? | ||
<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> | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | 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 <math>n</math> 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | For <math>x^2+x-n</math> to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals <math>\sqrt{1^2-4(1)(-n)}</math> or <math>\sqrt{4n+1}</math>. | ||
+ | |||
+ | Since <math>n</math> must be a positive integer, <math>4n+1</math> must be odd because if it were even, <math>4n</math> would have to be both odd and divisible by 4, which is a contradiction. Therefore, <math>n</math> must be even. | ||
+ | |||
+ | The maximum value of <math>4n+1</math> is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401. | ||
+ | There are 9: <math>9,25,49,81,121,169,225,289,</math>and <math>361</math>. Therefore, the answer is <math>\boxed{D}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1989|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:24, 27 June 2021
Contents
Problem
For how many integers between 1 and 100 does
factor into the product of two linear factors with integer coefficients?
Solution 1
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 ,
.
Solution 2
For to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals
or
.
Since must be a positive integer,
must be odd because if it were even,
would have to be both odd and divisible by 4, which is a contradiction. Therefore,
must be even.
The maximum value of is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401.
There are 9:
and
. Therefore, the answer is
.
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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