Question

How many ordered triples (A,B,C) of positive integers satisfy ABC = 4000?

Answer 1

Since 4000 = 2^5 5^3 we must have x = 2a^5d , y = 2b^5e and z = 2c^5f and
the answer will be AB , where A is the number of ordered triples (a, b, c)
of nonnegative integers such that a + b + c = 5 and B is the number of
ordered triples (d, e, f ) of nonnegative integers such that d + e + f = 3 .

Note that a = 0 has 6 possibilities for (b, c) , namely 0 ≤ b ≤ 5 , a = 1
has 5 possibilities for (b, c) , etc., down to a = 5 having one possible
(b, c) .
Thus, the number of possible (a, b, c) is 6 + 5 + 4 + 3 + 2 + 1 = 21 .
Similarly, the number of possible (d, e, f ) is 4 + 3 + 2 + 1 = 10 .
see whether it makes sense.