1987 AIME Problems/Problem 7

Problem

Let $[r,s]$ denote the least common multiple of positive integers $r$ and $s$ . Find the number of ordered triples $(a,b,c)$ of positive integers for which $[a,b] = 1000$ , $[b,c] = 2000$ , and $[c,a] = 2000$ .

Solution 1

It's clear that we must have $a = 2^j5^k$ , $b = 2^m 5^n$ and $c = 2^p5^q$ for some nonnegative integers $j, k, m, n, p, q$ . Dealing first with the powers of 2: from the given conditions, $\max(j, m) = 3$ , $\max(m, p) = \max(p, j) = 4$ . Thus we must have $p = 4$ and at least one of $m, j$ equal to 3. This gives 7 possible triples $(j, m, p)$ : $(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, 1, 4) , (3, 2, 4)$ and $(3, 0, 4)$ .

Now, for the powers of 5: we have $\max(k, n) = \max(n, q) = \max(q, k) = 4. Thus, at least two of$ k, n, q $must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples:$ (3, 3, 3) $and three possibilities of each of the forms$ (3, 3,3 n) $,$ (3, n, 13) $and$ (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

$1000 = 2^35^3$ and $2000 = 2^45^3$ . By looking at the prime factorization of $2000$ , $c$ must have a factor of $2^4$ . If $c$ has a factor of $5^3$ , then there are two cases: either (1) $a$ or $b = 5^32^3$ , or (2) one of $a$ and $b$ has a factor of $5^3$ and the other a factor of $2^3$ . For case 1, the other number will be in the form of $2^x5^y$ , so there are $4 \cdot 4 = 16$ possible such numbers; since this can be either $a$ or $b$ there are a total of $2(16)-1=31$ possibilities. For case 2, $a$ and $b$ are in the form of $2^35^x$ and $2^y5^3$ , with $x < 3$ and $y < 3$ (if they were equal to 3, it would overlap with case 1). Thus, there are $2(3 \cdot 3) = 18$ cases.

If $c$ does not have a factor of $5^3$ , then at least one of $a$ and $b$ must be $2^35^3$ , and both must have a factor of $5^3$ . Then, there are $4$ solutions possible just considering $a = 2^35^3$ , and a total of $4 \cdot 2 - 1 = 7$ possibilities. Multiplying by three, as $0 \le c \le 2$ , there are $7 \cdot 3 = 21$ . Together, that makes $31 + 18 + 21 = 70$ solutions for $(a, b, c)$ .

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1987 AIME Problems/Problem 7

Problem

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Solution 1

Solution 2

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