Trailing 0's

Which of the following can be the number of trailing zeros in n ! n! , where n n is a natural number ?

91 89 73 85

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1 solution

Maggie Miller
Aug 2, 2015

The number of trailing zeroes in 365 ! 365! is

365 5 + 365 25 + 365 125 = 73 + 14 + 2 = 89 \displaystyle\bigg\lfloor\frac{365}{5}\bigg\rfloor+\bigg\lfloor\frac{365}{25}\bigg\rfloor+\bigg\lfloor\frac{365}{125}\bigg\rfloor=73+14+2=\boxed{89} .

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