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

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

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TopNewestTrailing 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 \(z_n\). – Chew-Seong Cheong · 1 year, 8 months ago

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– Pi Han Goh · 1 year, 8 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 \(z_{8065} = 2012 + 1 = 2013 \) ,\( z_{8070} = 2013 + 1 =2014 < 2015\), \(z_{8075} = 2014 + 1 + 1 = 2016 > 2015 \).Log in to reply

– Chew-Seong Cheong · 1 year, 8 months ago

Thanks, Pi Han. I was using a spreadsheet, the table comes to me easy. I was showing how \(z_n\) changes with \(n\).Log in to reply

What is the proof of trailing zeroes? – Jåy Dïâz · 1 year ago

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trailing number of zeros – Pi Han Goh · 1 year ago

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