Proof problem about divisors

Question

Prove that number n = p₁^z1 p₂^z2 ... p_k^zk has (z₁+1)(z₂+1)...(z_n+1) divisors.

Answer 1

Hi,

n = p₁^z1 p₂^z2 ... p_k^zk has (z₁+1)(z₂+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

Answer 2

Well any divisor has to be of form p₁^a p₂^b .....

a can take values 0 ... z1, so (z1+1) ways

b can take values 0 ... z2 so (z2+1) ways

so the number of ways a,b ... can be chosen is (z1+1) (z2+2) ....