1992 AHSME Problems/Problem 13
Contents
Problem
How many pairs of positive integers (a,b) with satisfy the equation
Solution 1
We can rewrite the left-hand side as by multiplying the numerator and denominator by . Now we can multiply the top part of the fraction by the denominator to give . Now, as and are positive, we cannot have , so we must have , so the solutions are , which is 7 ordered pairs.
Solution 2
The equation is equivalent to or . The s cancel out leaving us with . This is asking us for the positive multiples of up to , and we know that , so our answer is .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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