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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.