Results 1 to 2 of 2

Math Help - divisible by 100

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    152

    divisible by 100

    Prove that if the difference of the integers a, b is divisible by 100, then
     a^{100}-b^{100}
    is divisible by 10 000.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,668
    Thanks
    298
    Awards
    1
    Quote Originally Posted by perash View Post
    Prove that if the difference of the integers a, b is divisible by 100, then
     a^{100}-b^{100}
    is divisible by 10 000.
    The best way to answer this depends on how much Math you've had.

    Here's one way:
    If (a - b)|100 then we can say that (a - b)k = 100, where k is a positive integer.

    Thus b = a  - \frac{100}{k}. So
    a^{100} - b^{100} = a^{100} - \left ( a - \frac{100}{k} \right )^{100}

    = a^{100} - \sum_{i = 0}^{100} (-1)^i{100 \choose i} a^{100 - i} \cdot \left ( \frac{100}{k} \right )^i

    = a^{100} - a^{100} - \sum_{i = 1}^{100} (-1)^i{100 \choose i} a^{100 - i} \cdot \left ( \frac{100}{k} \right )^i <-- Singling out the first term of the summation

    = -\sum_{i = 1}^{100} (-1)^i{100 \choose i} a^{100 - i} \cdot \left ( \frac{100}{k} \right )^i

    Now, each term in the series is divisible by at least 100 due to the factor of \frac{100}{k}. But there is another 100 involved: from the 100 \choose i, i > 0. So all terms are divisible by at least 10000.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 20th 2013, 09:32 AM
  2. Divisible by 9
    Posted in the Number Theory Forum
    Replies: 11
    Last Post: August 17th 2011, 02:48 AM
  3. Replies: 8
    Last Post: July 3rd 2011, 03:55 PM
  4. Replies: 1
    Last Post: May 7th 2010, 11:49 PM
  5. Replies: 5
    Last Post: January 1st 2010, 01:59 AM

Search Tags


/mathhelpforum @mathhelpforum