Question

Find the number of proper divisors of the number $N=2^p.3^q.5^r$. Also find the sum of
(i) all the divisors of N
(ii) all the proper divisors of N.
(iii) all the proper divisors of N which are multiple of 3.

Answer 1

Number of divisors is (p+1)(q+1)(r+1)
Number of proper divisors is
(p +1)(q +1)(r +1)- 2
where - 2 corresponds to 1 and N itself which are not proper divisors of N
Sum of all the divisors of N
${\sum }_p^{a=0}{2}^a{\sum }_q^{b=0}{3}^b{\sum }_{c=0}^r{5}^c$
$=\left(\frac{2^{p+1}-1}{2-1}\frac{3^{q+1}-1}{3-1}\frac{5^{r+1}-1}{5-1}\right)$ by sum of a G.P. ...(1)
Sum of all the proper divisors of N
Exclude 1 and N = $2^p{.3}^q{.5}^r$ from (1)
$\frac{1}{8}(2^{p+1}-1)(3^{q+1}-1)(5^{r+1}-1)-1-2^p{3}^q{5}^r$ ...(2)
If the divisor is to be a multiple of 3, then
$N=3(2^p.3^{q-1}.5^r)=3N^{\prime }$
Hence in this case the sum of all the divisors as in (1) will be
$3.\frac{1}{8}\left[\right(2^{p+1}-1\left)\right(3^q-1\left)\right(5^{r+1}-1\left)\right]$ ...(3)
Sum of all the proper divisors which are multiple of 3 will be obtained from (3) by excluding the number $2^p.3^q{.5}^r$ as it will not be proper.