# The product of all divisors of N is \[{{2}^{40}}{{3}^{10}}{{5}^{10}}\]. Find the sum of digits of N?

Answer

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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.

**Complete step by step 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.

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