Question

How many zeros should have 5 to have 72 dividers?

Answer 1

Hello,

nice as problem.


`image`
Here x has a power of 5 as divider and a power of 2 as divider (in order to make a power of 10.


`5=5^1*2^0`


`50=5^1*(5^1*2^1)=5^2*2^1`


`500=5^1*(5^2*2^2)=5^3*2^2`


`50 000 000=5^8*2^7`

Numbers of dividers: (8+1)*(7+1)=9*8=72