1987 AIME Problems/Problem 7
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Problem
Let 
denote the least common multiple of positive integers 
and 
. Find the number of ordered triples 
of positive integers for which 
, 
, and 
.
Contents |
Solution
Solution 1
It's clear that we must have 
, 
and 
for some nonnegative integers 
. Dealing first with the powers of 2: from the given conditions, 
, 
. Thus we must have 
and at least one of 
equal to 3. This gives 7 possible triples 
: 
and 
.
Now, for the powers of 5: we have 
. Thus, at least two of 
must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples: 
and three possibilities of each of the forms 
, 
and 
.
Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of 
possible valid triples.
Solution 2
and 
. By looking at the prime factorization of 
, 
must have a factor of 
. If has a factor of 
, then there are two cases: either (1) 
or 
, or (2) one of and 
has a factor of and the other a factor of 
. For case 1, the other number will be in the form of 
, so there are 
possible such numbers; since this can be either or there are a total of 
possibilities. For case 2, and are in the form of 
and 
, with 
and 
(if they were equal to 3, it would overlap with case 1). Thus, there are 
cases.
If does not have a factor of , then at least one of and must be 
, and both must have a factor of . Then, there are 
solutions possible just considering 
, and a total of 
possibilities. Multiplying by three, as 
, there are 
. Together, that makes 
solutions for 
.
See also
| 1987 AIME (Problems • Resources) | ||
| Preceded by Problem 6 | Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
