2007 AIME II Problems/Problem 2
Problem
Find the number of ordered triples where , , and are positive integers, is a factor of , is a factor of , and .
Solution
Denote and . The last condition reduces to . Therefore, is equal to one of the 9 factors of .
Subtracting the one, we see that . There are exactly ways to find pairs of if . Thus, there are solutions of .
Alternatively, note that the sum of the divisors of is (notice that after distributing, every divisor is accounted for). This evaluates to . Subtract for reasons noted above to get . Finally, this changes , so we have to add one to account for that. We get .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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