JEE Main & Advanced Mathematics Permutations and Combinations Number of Divisors

Let \[N=p_{1}^{{{\alpha }_{1}}}.p_{2}^{{{\alpha }_{2}}}.p_{3}^{{{\alpha }_{3}}}......p_{k}^{{{\alpha }_{k}}}\], where \[{{p}_{1}},\,{{p}_{2}},{{p}_{3}},......{{p}_{k}}\] are different primes and \[{{\alpha }_{1}},\,{{\alpha }_{2}},\,{{\alpha }_{3}},......,{{\alpha }_{k}}\] are natural numbers then :

(1) The total number of divisors of N including 1 and N is = \[({{\alpha }_{1}}+1)\,({{\alpha }_{2}}+1)\,({{\alpha }_{3}}+1)....({{\alpha }_{k}}+1)\]

(2) The total number of divisors of N excluding 1 and N is = \[({{\alpha }_{1}}+1)\,({{\alpha }_{2}}+1)\,({{\alpha }_{3}}+1).....({{\alpha }_{k}}+1)-2\].

(3) The total number of divisors of N excluding 1 or N is = \[({{\alpha }_{1}}+1)\,({{\alpha }_{2}}+1)\,({{\alpha }_{3}}+1).....({{\alpha }_{k}}+1)-1\].

(4) The sum of these divisors is    

\[=(p_{1}^{0}+p_{1}^{1}+p_{1}^{2}+......+p_{1}^{{{\alpha }_{1}}})\,(p_{2}^{0}+p_{2}^{1}+p_{2}^{2}+...+p_{2}^{{{\alpha }_{2}}}).....\]

\[(p_{k}^{0}+p_{k}^{1}+p_{k}^{2}+....+p_{k}^{{{\alpha }_{k}}})\]

(5) The number of ways in which N can be resolved as a product of two factors is

\[\left\{ \begin{matrix} \frac{1}{2}({{\alpha }_{1}}+1)\,({{\alpha }_{2}}+1)....({{\alpha }_{k}}+1),\,\text{If }N\text{ is not a perfect square}  \\ \frac{\text{1}}{2}[({{\alpha }_{1}}+1)\,({{\alpha }_{2}}+1).....({{\alpha }_{k}}+1)+1],\,\text{If }N\text{ is a perfect square}  \\ \end{matrix} \right.\]

(6) The number of ways in which a composite number \[N\] can be resolved into two factors which are relatively prime (or co-prime) to each other is equal to \[{{2}^{n-1}}\] where \[n\] is the number of different factors in \[N\].