Question

How many zeros should have 5 to have 72 dividers?

Answer 1

Hello,

nice as problem.

$| If\ a\ number\ x=a_1^{p_1}*a_2^{p_2}*a_3^{p_3}*....*a_n^{p_n}\\ the\ number\ of\ his\ dividers is:\\{p_1+1}*{p_2+1}*{p_3+1}*....*{p_n+1}$
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