Triple ABC

Triple ABC
How many ordered triples $\left (A,B,C\right )$ of positive integers satisfy $ABC = 4000$?
Source: mathcontest.olemiss.edu 9/10/2012

SOLUTION
This problem is similar to the problem “XYZ=4000” posted on 11/5/2014, nevertheless a different solution is offered here.

Let $N$ be a positive integer and $N=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ its prime factorization, then the number of divisors of $N$ equals $\left (e_1+1\right )\left (e_2+1\right )\cdots\left (e_k+1\right )$. For example,
$N=18=2^1\times 3^2$
Number of divisors
$\left (1+1\right )\left (2+1\right )=6$
Divisors
$1,2,3,6,9,18$

$N=4000=2^5\times 5^3$
Number of divisors
$\left (5+1\right )\left (3+1\right )=24$
Divisors
$R\!:1,2,4,8,16,32$
$S\!:5,10,20,40,80,160$
$T\!:25,50,100,200,400,800$
$U\!:125,250,500,1000,2000,4000$

We multiply the divisors to find the triples $\left (A,B,C\right )$ and  use multiplication tables to show the products.
$R\times U$ multiplication table

The $4000$ diagonal gives the $\left (1,B,C\right )$ triples; the $2000$ diagonal gives the $\left (2,B,C\right )$ triples; the $1000$ diagonal gives the $\left (4,B,C\right )$ triples, and so on. Beware that there will be duplicates. For simplicity, we list the triples in increasing numerical order.
$R\times U$ yields $12$ distinct triples
$\left (1,1,4000\right )\quad\left (2,2,1000\right )\quad\left (4,4,250\right )$
$\left (1,2,2000\right )\quad\left (2,4,500\right )\quad\;\;\left (4,8,125\right )$
$\left (1,4,1000\right )\quad\left (2,8,250\right )$
$\left (1,8,500\right )\quad\;\;\left (2,16,125\right )$
$\left (1,16,250\right )$
$\left (1,32,125\right )$

$S\times T$ multiplication table

$S\times T$ yields $21$ distinct triples
$\left (1,5,800\right )\quad\;\;\left (2,5,400\right )\quad\;\;\left (4,5,200\right )\quad\;\left (5,8,100\right )\quad\left (8,10,50\right )\quad\left (10,16,25\right )$
$\left (1,10,400\right )\quad\left (2,10,200\right )\quad\left (4,10,100\right )\quad\left (5,16,50\right )\quad\left (8,20,25\right )$
$\left (1,20,200\right )\quad\left (2,20,100\right )\quad\left (4,20,50\right )\quad\;\;\left (5,25,32\right )$
$\left (1,25,160\right )\quad\left (2,25,80\right )\quad\;\;\left (4,25,40\right )$
$\left (1,40,100\right )\quad\left (2,40,50\right )$
$\left (1,50,80\right )$

$R\times S$ yields duplicate triples
$R\times T$ yields duplicate triples
$S\times U$ yields no solution
$T\times U$ yields no solution
$R\times R$ yields no solution
$T\times T$ yields no solution
$U\times U$ yields no solution

$S\times S$ multiplication table

$S\times S$ yields $5$ distinct triples
$\left (5,5,160\right )\quad\left (10,10,40\right )$
$\left (5,10,80\right )\quad\left (10,20,20\right )$
$\left (5,20,40\right )$

The complete list of the $38$ distinct triples is shown below
The problem asks for ordered triples.
$\left (1,1,4000\right )\!,\left (2,2,1000\right )\!,\left (4,4,250\right )\!,\left (5,5,160\right )\!,\left (10,10,40\right )\!,\left (10,20,20\right )$ give
$6\times 3=18$ ordered triples.
The remaining $32$ triples give
$32\times 6=192$ ordered triples.
Total number of ordered triples $\left (A,B,C\right )$ such that $ABC=4000$
$18+192=210$

Answer: $210$