Trailing zeroes

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

Note by Joe Paradis
2 years, 11 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 \(z_n\).

Chew-Seong Cheong - 2 years, 11 months ago

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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 \(z_{8065} = 2012 + 1 = 2013 \) ,\( z_{8070} = 2013 + 1 =2014 < 2015\), \(z_{8075} = 2014 + 1 + 1 = 2016 > 2015 \).

Pi Han Goh - 2 years, 11 months ago

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

Chew-Seong Cheong - 2 years, 11 months ago

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What is the proof of trailing zeroes?

Jåy Dïâz - 2 years, 3 months ago

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Read trailing number of zeros

Pi Han Goh - 2 years, 3 months ago

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