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| In [[mathematics]], '''trailing zeros''' are a sequence of [[0 (number)|0]]s in the [[decimal]] representation (or more generally, in any [[positional notation|positional representation]]) of a number, after which no other [[Numerical digit|digit]]s follow.
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| Trailing zeros to the right of a [[decimal point]], as in 12.3400, do not affect the value of a number and may be omitted if all that is of interest is its numerical value. This is true even if the zeros [[recurring decimal|recur infinitely]]. However, trailing zeros may be useful for indicating the number of [[significant figure]]s, for example in a measurement. In such a context, "simplifying" a number by removing trailing zeros would be incorrect.
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| The number of trailing zeros in a base-''b'' [[integer]] ''n'' equals the exponent of the highest power of ''b'' that divides ''n''. For example, 14000 has three trailing zeros and is therefore divisible by 1000 = 10<sup>3</sup>. This property is useful when looking for small factors in [[integer factorization]]. [[Binary number]]s with many trailing zero bits are handled similarly in [[computer arithmetic]]. In computer software, the [[count trailing zeros]] operation efficiently determines the number of trailing zero bits in a machine word.
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| ==Factorial==
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| The number of trailing zeros in the [[decimal representation]] of ''n''!, the [[factorial]] of a [[non-negative]] [[integer]] ''n'', is simply the multiplicity of the [[prime number|prime]] factor 5 in ''n''<nowiki>!</nowiki>. This can be determined with this special case of [[de Polignac's formula]]:<ref>Summarized from [http://www.purplemath.com/modules/factzero.htm Factorials and Trailing Zeroes]</ref>
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| :<math>f(n) = \sum_{i=1}^k \left \lfloor \frac{n}{5^i} \right \rfloor =
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| \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, \,</math>
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| where ''k'' must be chosen such that | |
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| :<math>5^{k+1} > n \ge 5^k,\,</math>
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| and <math>\lfloor a \rfloor</math> denotes the [[floor function]] applied to ''a''. For ''n'' = 0, 1, 2, ... this is
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| :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, ... {{OEIS|A027868}}.
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| For example, 5<sup>3</sup> > 32, and therefore 32! = 263130836933693530167218012160000000 ends in
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| :<math>\left \lfloor \frac{32}{5} \right \rfloor + \left \lfloor \frac{32}{5^2} \right \rfloor = 6 + 1 = 7\,</math>
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| zeros. If ''n'' < 5, the inequality is satisfied by ''k'' = 0; in that case the sum is [[empty sum|empty]], giving the answer 0.
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| The formula actually counts the number of factors 5 in ''n''<nowiki>!</nowiki>, 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.
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| Defining
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| :<math>q_i = \left \lfloor \frac{n}{5^i} \right \rfloor,\,</math>
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| the following [[recurrence relation]] holds:
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| :<math>\begin{align}q_0\,\,\,\,\, & = \,\,\,n,\quad \\
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| q_{i+1} & = \left \lfloor \frac{q_i}{5} \right \rfloor.\,\end{align}</math>
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| This can be used to simplify the computation of the terms of the summation, which can be stopped as soon as ''q<sub> i</sub>'' reaches zero. The condition {{nowrap|5<sup>''k''+1</sup> > ''n''}} is equivalent to {{nowrap|1= ''q''<sub> ''k''+1</sub> = 0.}}
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| == See also ==
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| *[[Leading zero]]
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| ==References==
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| <references />
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| ==External links==
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| * [http://speleotrove.com/decimal/decifaq1.html#tzeros ''Why are trailing fractional zeros important?''] for some examples of when trailing zeros are significant
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| * [http://blog.dreamshire.com/2013/02/20/spoj-problem-11-factorial-fctrl/ ''Number of trailing zeros for any factorial''] Python program to calculate the number of trailing zeros for any factorial
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| [[Category:Elementary arithmetic]]
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| [[Category:Zero]]
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