<s>[INST] In how many ways can $7^{13}$ be represented as product of $3$ natural numbers? How i solved it:
all possible non-distinct groups $(a,b,c)$ are,
$a = 0 \Rightarrow (b,c) = (0,13)(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$
$a = 1 \Rightarrow (b,c) = (1,11)(2,10)(3,9)(4,8)(5,7)(6,6)$
$a = 2 \Rightarrow (b,c) = (2,9)(3,8)(4,7)(5,6)$
$a = 3 \Rightarrow (b,c) = (3,7)(4,6)(5,5)$
$a = 4 \Rightarrow (b,c) = (4,5)$
Thus, $7^{13}$ can be written as product of $3$ natural numbers in $7+6+4+3+1 = 21$ ways.
Though this gives the required solution, it takes time and is a lengthy way. Is there an alternate method to tackle such problems? (May be by using the "bars and stars" method with some adjustments? I tried but failed to get the correct answer that way.) Please help me out and share your method!
Thanks a lot!
 [/INST] Writing $7^{13}$ as a product of $3$ natural numbers becomes $7^{13}=7^{a}\times7^{b}\times7^{c}$
where $a,b,c$ are nonnegative integers with $a+b+c=13$. In fact
number $13$ must be split up in three parts. Do that by writing $15$
zero's on a row followed by picking out $2$ of them that are changed
into $1$. Then $13$ zeros are split up. There are $\left({15\atop 2}\right)=105$
choices when it comes to picking out $2$ from $15$.
Example: 
$000000000000000$ becomes $000100000100000$ that stands for possibility
$13=3+5+5$
$3$ on the left, $5$ in the middle and $5$ on the right.
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