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A problem goes like this: "If the product of the positive divisors of a number $N$ is: $2^{64} * 10^{48}$, find the sum of $N$'s digits." I don't know how to procede.


The only things I know are the formula for the amount of positive divisors:

For $N = p_1^a*p_2^b$. Then: $$\tau(n)=(a+1)(b+1)$$ And the formula for the product of the positive divisors (which I found on the Internet): $$\prod_{d|n}d = n^{\frac{\tau(n)}{2}}$$

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The product of the divisors of $N$ is $2^{112}5^{48}$ so all its divisors are of the form $2^{n}5^{m}$ indeed $N$ is of the form $2^{a}5^{b}$. The product of the divisors of $2^{a}5^{b}$ is \begin{eqnarray*} 2^{\frac{a(a+1)}{2}(b+1)} 5^{\frac{b(b+1)}{2}(a+1)} \end{eqnarray*} So we need to solve $\frac{a(a+1)}{2}(b+1)=112$ and $\frac{b(b+1)}{2}(a+1)=48$. So $a=7$ and $b=3$, giving $N=16000$. And the sum of the digits of this are $\color{red}{7}$.

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