Difference between revisions of "1987 AIME Problems/Problem 7"
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Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>. Thus, at least two of <math>k, n, q</math> 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: <math>(3, 3, 3)</math> and three possibilities of each of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>. | Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>. Thus, at least two of <math>k, n, q</math> 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: <math>(3, 3, 3)</math> and three possibilities of each of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>. | ||
− | Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of <math>7 \cdot 10 = | + | Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of <math>7 \cdot 10 = 70</math> possible valid triples. |
===Second solution=== | ===Second solution=== |
Revision as of 22:02, 6 March 2007
Problem
Let denote the least common multiple of positive integers and . Find the number of ordered triples of positive integers for which , , and .
Solution
First solution
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.
Second solution
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 .
c | a | b | solutions |
---|---|---|---|
31 | |||
18 | |||
24 |
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |