# The number of Zeros at the of100! is

Solution:

E2(100!)= The exponent of 2 in(100)!

$\\ \Rightarrow \left [ 100/2 \right ]+\left [ 100/2^2 \right ]+\left [ 100/2^3 \right ]+\left [ 100/2^4 \right ]+\left [ 100/2^5 \right ]+\left [ 100/2^6 \right ]\\ \\=50+25+12+6+3+1=97\\ \\\Rightarrow E5(100!)=\left [ 100/5 \right ]+\left [ 100/5^2 \right ]=20+4=24\\ \\\Rightarrow E10(100!)=Min{97,24}=24$

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