# Trailing zeroes

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

2 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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, 11 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, 11 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, 11 months ago

What is the proof of trailing zeroes?

- 2 years, 3 months ago