1989 AHSME Problems/Problem 8
Contents
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 , .
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 | |
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