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# Trailing zeroes

Can it be shown that no integer $$n$$ has an $$n!$$ with exactly 2015 trailing zeroes?

2 years, 6 months ago

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Trailing zeroes of $$n!$$ is given by $$z_n = \displaystyle \sum_{k=1}^\infty {\left \lfloor \frac{n}{5^k} \right \rfloor}$$, where $$\lfloor x \rfloor$$ is the greatest integer function. Therefore, there is a unique set of $$z_n$$. The values of $$z_n$$ around $$2015$$ are as follows:

$\begin{array} {} n: & 8055 & 8060 & 8065 & 8070 & 8075 & 8080 & 8085 & 8090 & 8095 \\ z_n: & 2011 & 2012 & 2013 & \color{red}{2014} & \color{red}{2016} & 2017 & 2018 & 2019 & 2020 \end{array}$

It is seen that $$2015$$ is not a $$z_n$$.

- 2 years, 6 months ago

There is no need to construct the table. You should show that the lower bound of $$n$$ satisfy the equation $$n\left(\frac15 +\frac1{5^2}+ \frac1{5^3}+\ldots\right) = 2015$$. Then you don't need to construct the table anymore and you're left to show that every increment by 5 of $$n$$ increases the trailing zeros by 1, and that the trailing zeros further increases by 1 when $$n$$ is a multiple of 25. Applying the formula, $$z_{8060} = 2012$$, so $$z_{8065} = 2012 + 1 = 2013$$ ,$$z_{8070} = 2013 + 1 =2014 < 2015$$, $$z_{8075} = 2014 + 1 + 1 = 2016 > 2015$$.

- 2 years, 6 months ago

Thanks, Pi Han. I was using a spreadsheet, the table comes to me easy. I was showing how $$z_n$$ changes with $$n$$.

- 2 years, 6 months ago

What is the proof of trailing zeroes?

- 1 year, 10 months ago