Question

The product of all divisors of N is $2^{40}{3}^{10}{5}^{10}$. Find the sum of digits of N.

Answer 1

Let N be a natural number number. Then it can be expressed 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.

Then number of divisors of N is given by :

$d=(a_1+1)(a_2+1).....(a_n+1)$

Now the product of divisors of a number N is given by $N^{\frac{d}{2}}$

Therefore by the problem we have,

$N^{\frac{d}{2}}=2^{40}{3}^{10}{5}^{10}$

$\to$ $N^{\frac{d}{2}}={16}^{10}{3}^{10}{5}^{10}$

$\to$ $N^{\frac{d}{2}}=(\mathrm{16.3.5}{)}^{10}$

$\to$ $N^{\frac{d}{2}}={240}^{10}$

$\to$$N^{\frac{d}{2}}={240}^{\frac{20}{2}}$

$\to$$N=240$ and $d=20$

(It can be cross verified that $240=2^4{3}^1{5}^1$ so $d=(4+1)(1+1)(1+1)=20$

Hence sum of the digits of $N=2+4+0=6$.