1987 AIME Problems/Problem 7

Problem

Now, for the powers of 5: we have $\max(k, n) = \max(n, q) = \max(q, k) = 3$ . 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 $(3, 3, n)$ , $(3, n, 3)$ and $(n, 3, 3)$ .

Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of $7 \cdot 10 = 070$ possible valid triples.

$1000 = 2^35^3$ 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 $(a, b, c)$ .