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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?

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Last updated date: 13th Jul 2024
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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.