Vishal Kataria

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Number of Divisors Part – II

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vishal

For any composite number n, where

prime factorization of n ={p_{1}}^{\alpha_{1}}\times{p_{2}}^{\alpha_{2}}\times...\times{p_{k}}^{\alpha_{k}}

the no. of divisors = \left ( \alpha_{1} +1\right )\times\left ( \alpha_{2} +1\right )\times...\times\left ( \alpha_{k} +1\right )

In this article, i am going to give the justification that why this formula works to calculate the number of divisors.

Lets start with an example:

Prime factorization of 12 = 2^{2}\times3^{1 }

Divisors of 2^{2} are 2^{0}, 2^{1} and 2^{2}  (total 3 in number i.e. 2+1)

Divisors of 3^{1}  are 3^{0}  and  3^{1}   (total 2 in number i.e. 1+1)

So, Divisors of 12 are 2^{0}.3^{0},  2^{0}.3^{1},  2^{1}.3^{0},  2^{1}.3^{1}, 2^{2}.3^{0}, 2^{2}.3^{1}   (total 3\times2 = 6 in number i.e.)

For any composite number ‘n’

Let the prime factorization of n be:

n ={p_{1}}^{\alpha_{1}}\times{p_{2}}^{\alpha_{2}}\times...\times{p_{k}}^{\alpha_{k}}

then no. of divisors of {p_{1}}^{\alpha_{1}} = \left ( \alpha_{1} +1\right )

then no. of divisors of {p_{2}}^{\alpha_{2}} = \left ( \alpha_{2} +1\right )

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then no. of divisors of {p_{k}}^{\alpha_{k}}   = \left ( \alpha_{k} +1\right )

then no. of divisors of n = \left ( \alpha_{1} +1\right )\times\left ( \alpha_{2} +1\right )\times...\times\left ( \alpha_{k} +1\right )

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