1987 AIME Problems/Problem 7

Problem

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

Solution

$\displaystyle 1000 = 2^35^3$ 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 $31 + 18 + 24 = 70$ solutions for $(a, b, c)$ .

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
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1987 AIME Problems/Problem 7

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