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Trailing zero
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<h2 id="factorial">Factorial</h2> <p>The number of trailing zeros in the <a href="/facts/Decimal_representation/HCLBt4UQ">decimal representation</a> of <i>n</i>!, the <a href="/facts/Factorial/kkfy1Bwe">factorial</a> of a <a href="/facts/Non-negative/z0xollg1">non-negative</a> <a href="/facts/Integer/PUwwrawx">integer</a> <i>n</i>, is simply the multiplicity of the <a href="/facts/Prime_number/L20FLkgn">prime</a> factor 5 in <i>n</i>!. This can be determined with this special case of <a href="/facts/De_Polignac%2527s_formula/paSjArnU">de Polignac's formula</a>:<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> f ( n ) = ∑ i = 1 k ⌊ n 5 i ⌋ = ⌊ n 5 ⌋ + ⌊ n 5 2 ⌋ + ⌊ n 5 3 ⌋ + ⋯ + ⌊ n 5 k ⌋ , {\displaystyle f(n)=\sum _{i=1}^{k}\left\lfloor {\frac {n}{5^{i}}}\right\rfloor =\left\lfloor {\frac {n}{5}}\right\rfloor +\left\lfloor {\frac {n}{5^{2}}}\right\rfloor +\left\lfloor {\frac {n}{5^{3}}}\right\rfloor +\cdots +\left\lfloor {\frac {n}{5^{k}}}\right\rfloor ,\,} <p>where <i>k</i> must be chosen such that </p> 5 k + 1 > n , {\displaystyle 5^{k+1}>n,\,} <p>more precisely </p> 5 k ≤ n < 5 k + 1 , {\displaystyle 5^{k}\leq n<5^{k+1},} k = ⌊ log 5 n ⌋ , {\displaystyle k=\left\lfloor \log _{5}n\right\rfloor ,} <p>and ⌊ a ⌋ {\displaystyle \lfloor a\rfloor } denotes the <a href="/facts/Floor_function/ssehyMMf">floor function</a> applied to <i>a</i>. For <i>n</i> = 0, 1, 2, ... this is </p> 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, ... (sequence A027868 in the <a href="/facts/On-Line_Encyclopedia_of_Integer_Sequences/yow6cVsU">OEIS</a>). <p>For example, 53 > 32, and therefore 32! = 263130836933693530167218012160000000 ends in </p> ⌊ 32 5 ⌋ + ⌊ 32 5 2 ⌋ = 6 + 1 = 7 {\displaystyle \left\lfloor {\frac {32}{5}}\right\rfloor +\left\lfloor {\frac {32}{5^{2}}}\right\rfloor =6+1=7\,} <p>zeros. If <i>n</i> < 5, the inequality is satisfied by <i>k</i> = 0; in that case the sum is <a href="/facts/Empty_sum/bvMsoUNh">empty</a>, giving the answer 0. </p><p>The formula actually counts the number of factors 5 in <i>n</i>!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. </p><p>Defining </p> q i = ⌊ n 5 i ⌋ , {\displaystyle q_{i}=\left\lfloor {\frac {n}{5^{i}}}\right\rfloor ,\,} <p>the following <a href="/facts/Recurrence_relation/JolXC71M">recurrence relation</a> holds: </p> q 0 = n , q i + 1 = ⌊ q i 5 ⌋ . {\displaystyle {\begin{aligned}q_{0}\,\,\,\,\,&=\,\,\,n,\quad \\q_{i+1}&=\left\lfloor {\frac {q_{i}}{5}}\right\rfloor .\,\end{aligned}}} <p>This can be used to simplify the computation of the terms of the summation, which can be stopped as soon as <i>q i</i> reaches zero. The condition 5<i>k</i>+1 > <i>n</i> is equivalent to <i>q</i> <i>k</i>+1 = 0. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Trailing_digit/RUfYtpgo">Trailing digit</a></li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="http://speleotrove.com/decimal/decifaq1.html#tzeros"><i>Why are trailing fractional zeros important?</i></a> for some examples of when trailing zeros are significant</li> <li><a href="http://blog.dreamshire.com/trailing_zeros_in_factorials/"><i>Number of trailing zeros for any factorial</i></a> Python program to calculate the number of trailing zeros for any factorial <a href="https://web.archive.org/web/20170222203959/http://blog.dreamshire.com/trailing_zeros_in_factorials/">Archived</a> 2017-02-22 at the <a href="/facts/Wayback_Machine/nmQ3a6JC">Wayback Machine</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Summarized from Factorials and Trailing Zeroes <a href="https://www.purplemath.com/modules/factzero.htm" target="_blank">https://www.purplemath.com/modules/factzero.htm</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>