Find the number of ways in which the number $6561^2$ can be written as a product of $3$ factors.

Question

Here is how I tried to solve this :-
$6561^2=3^{16}$
Number of factors for $3^{16} = 17$
Now $3^{16}=3^a\cdot3^b\cdot3^c$ in order to get $3^{16}$ as a product of $3$ numbers. Now
$a+b+c=16$ and we can use whole number distribution to get the number of ways which is $^{18}C_2 = 153$ number of ways but this is not the correct answer. What am I missing here? Please help me on this !!!

Is there any method to calculate the whole number distribution where all the numbers are distinct? I think I need to find the distinct powers of $3$. How many $3$ unique numbers can add up to 16 I guess that I should be looking for?

Thanks in advance !!!

Answer 1

(Based on your answer of 30,) What you want is the number of multi-sets $\{a, b, c\}$ whose product is $3^{16}$.
What this means is that $\{1, 1, 3^{16}\}, \{3^{16}, 1, 1\}$ should only be counted once. (Whereas when you do $a+b+c = 16$, you counted it 3 times.)

Hint: Essentially apply Polya Enumeration Theorem / Burnside lemma.

• Number of triples of the form $\{a, a, a\} - 0$.
• Number of triples of the form $\{a, a, b\} - 9$.
• Number of triples of the form $\{a, b, c\} - \frac{153 - 9\times 3}{6} = 21$
• Hence, total number of triples is $ 0 + 9 + 21 = 30$.

Now, I leave it to you to explain the above steps.