Question

Let n∈Z+and let
n = \[p _{1}^{k1}p _{2}^{k2}p _{3}^{k3} ... + p _{n}^{kn}\]

be its prime decomposition. How many divisors does n have?

Answer 1

i take that back, this is a choosing problem

Answer 2

for each of the k1 powers you can have k2 powers , etc
so it is
k1 * k2 * k3* ... kn

Answer 3

not quite

Answer 4

lets do an example
3^2 * 5^3

Answer 5

right, i need to add one

Answer 6

correct

Answer 7

the possibilities are for powers are
1,0
1,1
1,2
1,3

2,0
2,1
2,2
2,3

so we have 2 * (3+1)

Answer 8

for 3^2 * 5^3

Answer 9

0,0 is a possibility

Answer 10

thats a trivial divisor

Answer 11

still a divisor

Answer 12

ok,

Answer 13

0,0
0,1
0,2
0,3

Answer 14

right, just about to write that

Answer 15

so we have (2+1)(3+1). and generalizing this we have

(k1+1)(k2+1)...(kn+1) divisors

Answer 16

correct...\[\prod_{i=1}^{n}(k_i+1)\]

Answer 17

yes :)