Number of zeroes

Number Theory Level 3

Find the number of zeroes at the end of $9999!$ .

Clarification

For example, for $12! = 479,001,600$ the answer would be $2$ because if we look at the end there is a chain of $2$ zeroes

The answer is 2495.

3 solutions

Aditya Narayan Sharma
Mar 19, 2016

$E_5(9999!) = \lfloor \frac{9999}{5}\rfloor + \lfloor \frac{9999}{25}\rfloor + \lfloor \frac{9999}{125}\rfloor + \lfloor \frac{9999}{3125}\rfloor + \lfloor \frac{9999}{15625}\rfloor\ + \cdots$

$E_5(9999!) = 1999 + 399 + 79 + 15 + 3 + 0 = \boxed{2495}$

For the first addend in the second equation, did you mean to write $1999$ , not $2399$ ?

Ralph James - 5 years, 2 months ago

Yeah I'm sorry typing mistake. I've fixed that now

Aditya Narayan Sharma - 5 years, 2 months ago

Arulx Z
Mar 24, 2016

n = int(raw_input())
c = 5

r = n // c

count = 0

while r > 0:
    count += r
    c *= 5
    r = n // c

print count

Tom Engelsman
Nov 26, 2020

The number of trailing zeros in $9999!$ is computed per:

$\lfloor \frac{9999}{5} \rfloor =1999 \Rightarrow \lfloor \frac{1999}{5} \rfloor = 399 \Rightarrow \lfloor \frac{399}{5} \rfloor = 79 \Rightarrow \lfloor \frac{79}{5} \rfloor = 15 \Rightarrow \lfloor \frac{15}{5} \rfloor = 3 < 5$

Thus, we have $1999 +399 + 79 + 15 + 3 = \boxed{2495}$ trailing zeros.

Number of zeroes

The answer is 2495.

3 solutions

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