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 
.
Video Solution by OmegaLearn
https://youtu.be/LqrXinQbk1Q?t=73
~ pi_is_3.14
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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