XYZ=4000

How many ordered triples $(x,y,z)$ of positive integers satisfy $xyz=4000$?
Source: NCTM Mathematics Teacher 2006

SOLUTION
This problem is similar to the problem “Triple ABC” posted on 9/14/2012, nevertheless a different solution is offered here.

Given that $4000=2^5\!\cdot 5^3$, the positive integers $x,y,z$ must take the form $x=2^a\!\cdot 5^d$, $y=2^b\!\cdot 5^e$, and $z=2^c\!\cdot 5^f$ where $a,b,c,d,e,f$ are non-negative integers.
$xyz=2^a\!\cdot 2^b\!\cdot 2^c\!\cdot 5^e\!\cdot 5^e\!\cdot 5^f$
$=2^{a+b+c}\!\cdot 5^{d+e+f}$
The more difficult problem of finding the single product $xyz=4000$ turns into the much simpler problem of finding two products, the first being $2^a\!\cdot 2^b\!\cdot 2^c$ such that $a+b+c=5$ and the second $5^d\!\cdot 5^e\!\cdot 5^f$ such that $d+e+f=3$.
If $a=0, a+b+c$ have $6$ possible values
$0+0+5$
$0+1+4$
$0+2+3$
$0+3+2$
$0+4+1$
$0+5+0$
If $a=1, a+b+c$ have $5$ possible values
$1+0+4$
$1+1+3$
$1+2+2$
$1+3+1$
$1+4+0$
If $a=2, a+b+c$ have $4$ possible values
$2+0+3$
$2+1+2$
$2+2+1$
$2+3+0$
If $a=3, a+b+c$ have $3$ possible values
If $a=4, a+b+c$ have $2$ possible values
If $a=5, a+b+c$ have $1$ possible value
Total = $6+5+4+3+2+1=21$ possible ways to choose $2^a\!\cdot 2^b\!\cdot 2^c$
Similarly, we want $d+e+f=3$.
If $d=0, d+e+f$ have $4$ possible values
If $d=1, d+e+f$ have $3$ possible values
If $d=2, d+e+f$ have $2$ possible values
If $d=3, d+e+f$ have $1$ possible value
Total = $4+3+2+1=10$ possible ways to choose $5^d\!\cdot 5^e\!\cdot 5^f$
There are $21\cdot 10=210$ ways to choose $2^a\!\cdot 2^b\!\cdot 2^c\!\cdot 5^d\!\cdot 5^e\!\cdot 5^f$; there are $210$ ordered triples $(x,y,z)$ such that $xyz=4000$.

Answer: $210$