12.0 Total Number of Possible Divisors for a Given Natural Number

Let '$n$' be a natural number. It can be given as product of prime factors.

Consider ${p_1}$ , ${p_2}$ , ${p_3}$ and so on to be distinct prime numbers. Thus $n$ can be written as,

$$ \Rightarrow n = {p_1}^{{\alpha _1}}.{p_2}^{{\alpha _2}}.{p_3}^{{\alpha _3}}.{p_4}^{{\alpha _4}}.......$$

where ${\alpha _k}$ is always a whole number.

Let $m$ be a divisor of $n$.

Thus, $$ \Rightarrow m = {p_1}^{{\beta _1}}.{p_2}^{{\beta _2}}.{p_4}^{{\beta _4}}.......$$

The number of factors is equal to the number of ways of selecting ${\beta _k}$ which is less than or equal to ${\alpha _k}$.i.e.

$${\beta _k} \leqslant {\alpha _k}$$

Then the total number of such divisors (factors) is given by,

$$ = ({\alpha _1} + 1)({\alpha _2} + 1)({\alpha _3} + 1).......$$

Note: When proper factors are to be selected, one and the number itself are to be excluded, i.e. selecting none and selecting all at a time are to be excluded.

Then the total number of such divisors (factors) are, $$ = [({\alpha _1} + 1)({\alpha _2} + 1)({\alpha _3} + 1).......]-1-1$$ $$\begin{equation} \begin{aligned} = [({\alpha _1} + 1)({\alpha _2} + 1)({\alpha _3} + 1).......] - 1 - 1 \\ = [({\alpha _1} + 1)({\alpha _2} + 1)({\alpha _3} + 1).......] - 2 \\\end{aligned} \end{equation} $$