Vishal Kataria

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Number of Divisors

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How to find Number of Divisors of a Composite Number?

Composite Numbers: Number (greater than one) which has more than two factors is known as a composite number.

Is there any method to know the ‘Number of Divisors’ of a composite number without actually finding the divisors of a number?

Answer is yes!

Before that we need to know ‘The Fundamental Theorem of Arithmetic’ & Prime Factorization.

Fundamental Theorem of Arithmetic: Every natural number greater than 1 can uniquely be expressed as product of primes except the order of the factors.

FORMULA 

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 n = \left ( \alpha_{1} +1\right )+\left ( \alpha_{2} +1\right )+...+\left ( \alpha_{k} +1\right )

Examples:

Divisors of 4

Prime factorization of 4 =  2^{2}

No. of divisors of 4  = (2+1)  = 3

Divisors of 12

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

No. of divisors of 12 = (2+1)(1+1) = 6

Divisors of 90

Prime factorization of 90 =  2^{1}\times3^{2}\times5^{1}

No. of divisors of 90 = (1+1)(2+1)(1+1) = 12

Divisors of 5600

Prime factorization of 5600= 2^{5}\times5^{2}\times7^{1}

No. of divisors of 5600 = (5+1)(2+1)(1+1)

PROOF

For proof or generalization of this formula click on PROOF

for 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 )+\left ( \alpha_{2} +1\right )+...+\left ( \alpha_{k} +1\right )

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