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# The product of all divisors of N is ${{2}^{40}}{{3}^{10}}{{5}^{10}}$. Find the sum of digits of N? Verified
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Hint: In this problem, we have to find the product of all divisors of N is ${{2}^{40}}{{3}^{10}}{{5}^{10}}$and the sum of digits of N. We can first find the product of all divisors of N using the format ${{N}^{\dfrac{d}{2}}}$, where
$d=\left( {{a}_{1}}+1 \right)\left( {{a}_{2}}+1 \right)......\left( {{a}_{n}}+1 \right)$ is the number of divisors of N, here ${{a}_{1}},{{a}_{2}},..$are the power terms of the given numbers. We can then add the digits in the product of divisors, to get the answer.

Here we have to find product of all divisors of N is ${{2}^{40}}{{3}^{10}}{{5}^{10}}$and the sum of digits of N.
We can assume that,
Let N be natural number, we can now express it in the form,
$p_{1}^{{{a}_{1}}},p_{2}^{{{a}_{2}}},....p_{n}^{{{a}_{n}}}$
Where, ${{p}_{1}},{{p}_{2}},...{{p}_{n}}$ are prime numbers.
We know that, the number of divisors of N can be written as,
$d=\left( {{a}_{1}}+1 \right)\left( {{a}_{2}}+1 \right)......\left( {{a}_{n}}+1 \right)$…… (1)
Now, we can write the product of divisors of a number N in the form ${{N}^{\dfrac{d}{2}}}$, we get
$\Rightarrow {{N}^{\dfrac{d}{2}}}={{2}^{40}}{{3}^{10}}{{5}^{10}}$
We can now write it as,
$\Rightarrow {{N}^{\dfrac{d}{2}}}={{16}^{10}}{{3}^{10}}{{5}^{10}}$
We can now simplify the above step, we get
\begin{align} & \Rightarrow {{N}^{\dfrac{d}{2}}}={{\left( 16\times 3\times 5 \right)}^{10}} \\ & \Rightarrow {{N}^{\dfrac{d}{2}}}={{240}^{10}} \\ & \Rightarrow {{N}^{\dfrac{d}{2}}}={{240}^{\dfrac{20}{2}}} \\ \end{align}
We can now verify whether 240 is an equivalent to the power terms,
$\Rightarrow 240={{2}^{4}}{{3}^{1}}{{5}^{1}}$
We can now substitute these power terms in (1), we get
$d=\left( 4+1 \right)\left( 1+1 \right)+\left( 1+1 \right)=20$
We can now add the N digits, we get
$\Rightarrow N=2+4+0=6$
Therefore, if the product of all divisors of N is ${{2}^{40}}{{3}^{10}}{{5}^{10}}$, then the sum of digits of N is 6.

Note: We should always remember that the number of divisors of N can be written as,
$d=\left( {{a}_{1}}+1 \right)\left( {{a}_{2}}+1 \right)......\left( {{a}_{n}}+1 \right)$ and we can write the product of divisors of a number N in the form ${{N}^{\dfrac{d}{2}}}$, where ${{a}_{1}},{{a}_{2}},..$are the power terms of the given numbers.