Exercise 4.1.2 (Number of zeros in a row at the right end of $100!$)

Question

If $100!$ were written out in the ordinary decimal notation without the factorial sign, how many zeros would there be in a row at the right end?

Richard Ganaye · Accepted Answer

\begin{proof} 
The number $N$ of zeros in a row at the right end of the decimal expansion of $100!$ is the highest power of $10$ dividing $100!$. As in Problem 1,
\begin{align*}
N & = \min(
u_p(2), 
u_p(5))\
&= 
u_5(100!).\
\end{align*}
The Legendre-de Polignac's algorithm gives
$$
\begin{array}{lll}
\lfloor 100/ 5 floor = 20, & \lfloor 20 / 5 floor = 4, & \lfloor 4 / 5 floor = 0.
\end{array}
$$
Thus $N = 
u_p(100!) = 24$.
\end{proof}

Check: 
\begin{verbatim}
sage: n = factorial(100); n
933262154439441526816992388562667004907159682643816214685929638952175999932299156
08941463976156518286253697920827223758251185210916864000000000000000000000000

sage: counter = 0
sage: while n % 10 == 0:
....:     n = n // 10
....:     counter += 1
....: print(counter)
           24
\end{verbatim}

Exercise 4.1.2 (Number of zeros in a row at the right end of $100!$)

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Exercise 4.1.2 (Number of zeros in a row at the right end of $100!$)

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