Results 1 to 6 of 6

Math Help - factorial

  1. #1
    Newbie
    Joined
    Aug 2010
    From
    iligan.philippines
    Posts
    10

    factorial

    There are k zeros at the end of 34! = 34 x 33 x  32 x .... 4 x 3 x 2 x 1: What
    is the value of k?

    I know the answer is 7 zeros but I can't show it with a complete solution. Can anybody help me with this,please...........
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Consider the prime factorization of 34!. Since 2's are in abundance, we only need to calculate the number of 5's. 5, 10, 15, 20 and 30 yield one 5 each while 25 yields two 5's. So there are seven 5's in the prime factorization of 34!. Hence there are seven zeros at the end of 34!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    83
    Thanks
    1
    no. of zeros means no. of tens (10)

    to get 10, two factors are required 2 and 5.(10=2X5)

    so it is enough to find how many 2's and 5's are there.

    1. find no. of 2's

    34/2=17
    17/2=8
    8/2=4
    4/2=2
    2/2=1

    do the successive division and add all quotient, we get no. of 2's =17+8+4+2+1=32.

    2. find no. of 5's

    34/5=6
    6/5=1

    no. of 5's =6+1=7

    so 34!=(some numbers) X 2^32 X 5^7.

    therefore no of zeros or tens is maximum no of combination of 2 &
    maximum 7 combination is possible.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig  ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{34}  {5^i}\right\rfloor=\left\lfloor\frac{34}{5}\right\  rfloor+\left\lfloor\frac{34}{5^2}\right\rfloor=7

    At a certain point, the values will all be zero.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1
    Quote Originally Posted by dwsmith View Post
    \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig  ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{34}  {5^i}\right\rfloor=\left\lfloor\frac{34}{5}\right\  rfloor+\left\lfloor\frac{34}{5^2}\right\rfloor=7

    At a certain point, the values will all be zero.
    here [.] indicates greatest integer funtion right??

    and is it only applicable for finding prime factors??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    That notation is the floor and the formula finds the number of trailing zeros.

    For instance, let's look at numbers we can easily see 5! = 120 and 10! = 3628800. We have 1 and 2 trailing zeros, respectively.

    \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig  ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{5}{  5^i}\right\rfloor=\left\lfloor\frac{5}{5}\right\rf  loor=1

    \displaystyle k=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{5^i}\rig  ht\rfloor=\sum_{i=1}^{\infty}\left\lfloor\frac{10}  {5^i}\right\rfloor=\left\lfloor\frac{10}{5}\right\  rfloor=2
    Last edited by dwsmith; January 8th 2011 at 08:06 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum of Factorial
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 4th 2011, 12:10 PM
  2. Factorial N
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 28th 2010, 10:18 PM
  3. factorial!
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 18th 2010, 03:58 PM
  4. limt with factorial
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 9th 2009, 08:39 PM
  5. factorial
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 15th 2009, 07:13 AM

Search Tags


/mathhelpforum @mathhelpforum