Question

The number of positive divisors of ${2^5}{3^6}{7^3}$ is
$\eqalign{
& A)\,\,\,14 \cr
& B)\,\,\,167 \cr
& C)\,\,\,168 \cr
& D)\,\,210 \cr} $

Answer 1

Hint:
If N be an integer $ > 1\,\,\,and\,\,N = {a^p} \times {b^q} \times {c^r} \times ...$ where a,b,c,… are prime factors of N and p,q,r ,…are natural powers of a,b,c,…respectively.Then the number of positive divisors of N is given by $$d = \left( {p + 1} \right)\left( {q + 1} \right)\left( {r + 1} \right) \ldots $$ We have to split any number into that form to find number of positive divisors.

Complete step by step solution:
Step1:
   Here the given number $N = {2^5}{3^6}{7^3}$ .
Step2:
  Also a=2, b=3, c=7 are prime factors of N.
Step3:
  p=5, q=6, r=3 are natural powers.
Step4:
 Then the number of positive divisors
                    $\eqalign{
  & d = (p + 1)(q + 1)(r + 1) \cr
  & \,\,\,\,\, = (5 + 1)(6 + 1)(3 + 1) \cr
  & \,\,\,\,\, = 6 \times 7 \times 4 \cr
  & \,\,\,\,\, = 168 \cr} $

Hence, the option C) is correct.

Note:
From where does the formula come? Let us take an example of small number 48.Now $48 = {2^4} \times 3$ .Here p=4,q=1.Here 2 and 3 are prime factors of N=48.Here ${2^0}, {2^1}, {2^2}, {2^3}, {2^4}$ all are positive divisors of 48,where power $2^0$ means no occurrence of 2. And $2^{m}$ means m number of 2’s occur in a divisor of 48. Then 2 comes as a divisor in (4+1) ways. And in the each way 3 come as a divisor in (1+1) ways.Hence the total number of positive divisors
$\eqalign{
  & d = (4 + 1)(1 + 1) \cr
  & = 10 \cr
  & = (p + 1)(q + 1) \cr} $
Hence,the formula holds.