Difference between revisions of "1987 AIME Problems/Problem 7"
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== Problem == | == Problem == | ||
Let <math>[r,s]</math> denote the [[least common multiple]] of [[positive integer]]s <math>r</math> and <math>s</math>. Find the number of [[ordered tuple | ordered triples]] <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. | Let <math>[r,s]</math> denote the [[least common multiple]] of [[positive integer]]s <math>r</math> and <math>s</math>. Find the number of [[ordered tuple | ordered triples]] <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. | ||
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== Solution == | == Solution == | ||
===Solution 1=== | ===Solution 1=== |
Revision as of 22:12, 23 November 2007
Contents
[hide]Problem
Let denote the least common multiple of positive integers and . Find the number of ordered triples of positive integers for which , , and .
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 • 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 |