Difference between revisions of "1987 AIME Problems/Problem 7"
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==Solution 1== | ==Solution 1== | ||
− | It's clear that we must have <math>a = 2^j5^k</math>, <math>b = 2^m 5^n</math> and <math>c = 2^p5^q</math> for some [[nonnegative]] [[integer]]s <math>j, k, m, n, p, q</math>. Dealing first with the powers of 2: from the given conditions, <math>\max(j, m) = 3</math>, <math>\max(m, p) = \max(p, j) = 4</math>. Thus we must have <math>p = 4</math> and at least one of <math>m, j</math> equal to 3. This gives 7 possible triples <math>(j, m, p)</math>: <math>(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, | + | It's clear that we must have <math>a = 2^j5^k</math>, <math>b = 2^m 5^n</math> and <math>c = 2^p5^q</math> for some [[nonnegative]] [[integer]]s <math>j, k, m, n, p, q</math>. Dealing first with the powers of 2: from the given conditions, <math>\max(j, m) = 3</math>, <math>\max(m, p) = \max(p, j) = 4</math>. Thus we must have <math>p = 4</math> and at least one of <math>m, j</math> equal to 3. This gives 7 possible triples <math>(j, m, p)</math>: <math>(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, 1, 4) , (3, 2, 4)</math> and <math>(3, 0, 4)</math>. |
− | Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = | + | Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 4. 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, 3,3 n)<math>, </math>(3, n, 13)<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 = 70 | + | Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of </math>7 \cdot 10 = 70$ possible valid triples. |
==Solution 2== | ==Solution 2== |
Revision as of 09:00, 18 June 2024
Problem
Let denote the least common multiple of positive integers and . Find the number of ordered triples of positive integers for which , , and .
Contents
[hide]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 k, n, q(3, 3, 3)(3, 3,3 n)(3, n, 13)(n, 3, 3)$.
Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of$ (Error compiling LaTeX. Unknown error_msg)7 \cdot 10 = 70$ 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 |
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