Touba M. answered • 12/29/20

B.S. in Mathematics with 20+ years of Teaching and Tutoring Experience

Hi,

n = p_{1}^{z1 }p_{2}^{z2 }... ^{ }p_{k}^{zk }has (z_{1}+1)(z_{2}+1)...(z_{n}+1) divisors.

First of all, let me make an example then prove it

n = 12 = 2 ^**2** * 3^**1** so the number of divisor will be (**2** + 1 ) ( **1** + 1 ) = 6

you know all divisors will be { 1, 2, 3, 4, 6, 12} that you see the number is 6.

Now we need to prove

The first exponent is z1 in fact P1 ^ z1 includes p1 ^ 0, p1 ^ 1, p1 ^ 2 , and so on till to p1 ^z1 it means the number of p1 ^ z1 includes z1 + 1 because started of 0

the same reason p2 ^ z2 will be z2 + 2

as a result the number of divisor is ( z1 + 1 ) ( z2 + 2) .... ( zn + 1)

I hope it is useful,

Minoo