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Math Help - on findin nos of 0's at end of a factorial

  1. #1
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    on findin nos of 0's at end of a factorial

    find 0's at end of 1*2^2 *3^3 *......100^100??
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  2. #2
    Super Member PaulRS's Avatar
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    Let <br />
v_p \left( n \right)<br />
be the maximum <br />
s \in \mathbb{N}<br />
such that <br />
\left. {p^s } \right|\left( {1^1  \cdot 2^2  \cdot ... \cdot n^n } \right)<br />

    First note that <br />
v_5 \left( n \right) \leqslant v_2 \left( n \right)<br />
so the number of 0s at the end of ^1  \cdot 2^2  \cdot ... \cdot n^n will be exactly <br />
v_5 \left( n \right)<br />



    For example for n=25 we have  v_5 \left( n \right)=5+10+15+20+2\cdot{25}, note that the coefficients of the multiples of 5 are the maximum powers of 5 dividing these numbers. This comes from <br />
\left( {5^a  \cdot k} \right)^{5^a  \cdot k}  = 5^{\boxed{a \cdot 5^a  \cdot k}}  \cdot k^{a \cdot 5^a  \cdot k} <br />
- (k,5)=1- So the maximum power of 5 dividing the number a^a is itself a multiplied by the maximum s such that 5^s|a

    Now since <br />
\left\lfloor {\tfrac{j}<br />
{n}} \right\rfloor  - \left\lfloor {\tfrac{{j - 1}}<br />
{n}} \right\rfloor  = \left\{ \begin{gathered}<br />
  1{\text{ if }}\left. n \right|j \hfill \\<br />
  0{\text{ otherwise}} \hfill \\ <br />
\end{gathered}  \right.<br />
it follows that the maximum s such that 5^s|j is <br />
{\sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{j}<br />
{{5^k }}} \right\rfloor  - \left\lfloor {\tfrac{{j - 1}}<br />
{{5^k }}} \right\rfloor } }<br />

    Thus: <br />
v_p \left( n \right) = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{j}<br />
{{5^k }}} \right\rfloor  - \left\lfloor {\tfrac{{j - 1}}<br />
{{5^k }}} \right\rfloor } } \right) \cdot j} <br />

    It may also be written as: <br />
\sum\limits_{i = 1}^{\left\lfloor {\frac{n}<br />
{5}} \right\rfloor } {5 \cdot i}  + \sum\limits_{i = 1}^{\left\lfloor {\frac{n}<br />
{{25}}} \right\rfloor } {25 \cdot i}  + ... + \sum\limits_{i = 1}^{\left\lfloor {\frac{n}<br />
{{5^{\left\lfloor {\log _5 \left( n \right)} \right\rfloor } }}} \right\rfloor } {5^{\left\lfloor {\log _5 \left( n \right)} \right\rfloor }  \cdot i} <br />
( note that a appears in as many sums as the maximum s such that 5^s|a)

    PS- That's not a factorial
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  3. #3
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    Hello, chandni!

    Opalg pointed out a big error in my calculatitons . . . *blush*


    Find the number of 0's at the end of: . 1^1\cdot2^2\cdot3^3\cdot4^4\cdots100^{100}

    Every factor of 5 (combined with an even number) produces a final zero.

    How many 5's are in that product?


    Zeros arise from: . 5^5\cdot 10^{10} \cdot 15^{15} \cdot 20^{20} \cdots 100^{100} \;= \;(5^5)\,(2^{10}5^{10})\,(3^{15}5^{15})\,(4^{20}5^  {20}) \cdots (20^{100}5^{100})


    We are concerned with the factors of 5 (only).
    \text{So we have: }\;5^5\cdot5^{10}\cdot5^{15}\cdot5^{20} \cdots 5^{100}  \;=\;5^{\overbrace{(5+10+15+20+\cdots+100)}^{\text  {arithmetic series}}}


    The sum of the series is 1050
    . . Hence, the product contains a factor of 5^{1050} . . . . no


    I overlooked these factors: . {\color{blue}25^{25},\;50^{50},\;75^{75},\;100^{10  0}}

    . . {\color{blue}\begin{array}{cccccc}25^{25} &=& (5^2)^{25} & =& 5^{50} & \text{50 zeros, not 25} \\<br />
50^{50} &=& 2^{50}(5^2)^{50} &=& 2^{50}5^{100} & \text{100 zeros, not 50}\\<br />
75^{75} &=& 3^{75}(5^2)^{75} &=& 3^{75}5^{150} & \text{150 zeros, not 75} \\<br />
100^{100} &=& 4^{100}(5^2)^{100} &=& 4^{100}5^{200}  & \text{200 zeros, not 100}\end{array}}

    My total was short by: . {\color{blue}25 + 50 + 75 + 100 \:=\:250} fives.

    The product contains a factor of: . {\color{blue}5}^{{\color{red}1300}}

    Therefore, the product ends in 1300 zeros.


    Thank you, Opalg!
    .
    Last edited by Soroban; November 15th 2008 at 02:43 PM.
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  4. #4
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    recheck ur solution

    hey i got dat but m confused as to y u included 100^100 and 50^50 in ur calculations the 2nd time wen u had already included it previously...jus chk out dat n do reply
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