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Thread: 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 $\displaystyle
    v_p \left( n \right)
    $ be the maximum $\displaystyle
    s \in \mathbb{N}
    $ such that $\displaystyle
    \left. {p^s } \right|\left( {1^1 \cdot 2^2 \cdot ... \cdot n^n } \right)
    $

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



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

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

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

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

    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: .$\displaystyle 1^1\cdot2^2\cdot3^3\cdot4^4\cdots100^{100}$

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

    How many 5's are in that product?


    Zeros arise from: .$\displaystyle 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).
    $\displaystyle \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 $\displaystyle 1050$
    . . Hence, the product contains a factor of $\displaystyle 5^{1050}$ . . . . no


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

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

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

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

    Therefore, the product ends in 1300 zeros.


    Thank you, Opalg!
    .
    Last edited by Soroban; Nov 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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