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lucianacp77 lucianacp77
  • 19-05-2016
  • Mathematics
Answered

how many zeros should have 5 to have 72 dividers?

Verified Answer :

caylus
caylus caylus
  • 20-05-2016

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)=98=72








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