How many ordered triples $(a, b, c)$ of positive integers exist with the property that $abc = 500$?
Breaking it up, $500 = 2^2\cdot5^3$
$abc = 2^2 \cdot 5^3 = 2\cdot 2 \cdot 5 \cdot 5 \cdot 5$
But how can I use this?
How many ordered triples $(a, b, c)$ of positive integers exist with the property that $abc = 500$?
Breaking it up, $500 = 2^2\cdot5^3$
$abc = 2^2 \cdot 5^3 = 2\cdot 2 \cdot 5 \cdot 5 \cdot 5$
But how can I use this?
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We solve a slightly more general problem, the number of triples $(a,b,c)$ of positive integers such that $abc=n=2^s5^t$. The solution generalizes nicely to $k$-tuples, and any positive $n$, given the prime power factorization of $n$.
The answer is the number of ways to distribute the $s$ $2$'s between $a$, $b$, and $c$, times the number of ways to distribute the $t$ $5$'s between $a$, $b$, and $c$.
By Stars and Bars (please see Wikipedia) there are $\binom{s+3-1}{3-1}$ ways to distribute the $2$'s, and $\binom{t+3-1}{3-1}$ ways to distribute the $5$'s.
In our particular case we get $\binom{4}{2}\binom{5}{2}$.
Remark: The solution is overkill. For $n=500$, we can count explicitly the ways to distribute the $2$'s, and the ways to distribute the $5$'s. Let's do the $5$'s. They all could go to one person ($3$ ways). Or else they could be distributed evenly ($1$ way). Or $2$ could go to one person, and $1$ to another ($(3)(2)$ ways). So there are $10$ ways to distribute the $5$'a. A similar but simpler analysis shows there are $6$ ways to distribute the $2$'s.