Results 1 to 5 of 5

Math Help - Finding the difference between last 2 digits of 7^2010

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    95

    Finding the difference between last 2 digits of 7^2010

    How to go about this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle 7^{2010}\equiv x \ \mbox{(mod 100)}

    \displaystyle 7^5\equiv 07 \ \mbox{(mod 100)}

    \displaystyle (7^{5})^{402}\equiv 7^{402}\equiv 7^{400+2}\equiv 7^{400}*7^2\equiv (7^5)^{80}*49\equi 7^{80}*49 \ \mbox{(mod 100)}

    Now as Soroban said 7^4\equiv 1 \ \mbox{mod 100}

    \displaystyle 7^{80}*49\equiv (7^4)^{20}*49\equiv 1^{20}*49\equiv 49  \ \mbox{(mod 100)}

    Sorry about the mistake
    Last edited by dwsmith; December 11th 2010 at 10:16 AM. Reason: corrected due to error
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    95
    thanks a lot
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by dwsmith View Post
    \displaystyle 7^{2010}\equiv x \ \mbox{(mod 100)}

    \displaystyle 7^5\equiv 07 \ \mbox{(mod 100)}

    \displaystyle (7^{402})^{5}\equiv 07 \ \mbox{(mod 100)}

    I'm not sure I get the above: the powers of 7 have a periodicy length of 4, and all the powers of 7 which are 2 mod 4 end with 49, so

    the difference must be 9-4 = 5 ...Is this what you meant?

    Also, why to write 7^5 = 07\!\!\pmod {100} and not simply 7^5=7\!\!\pmod {100} ?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742
    Hello, pranay!

    \text{Find the difference between the last two digits of }7^{2010}

    We have: . 7^4 \;=\; 2401 \quad\Rightarrow\quad 7^4 \;\equiv\; 1\text{ (mod 100)}

    Then: . 7^{2010} \;=\; 7^{4(502) + 2} \;=\; 7^{4(502)}\cdot7^2 \;=\;(7^4)^{502}\cdot49

    Hence: . (7^4)^{502}\cdot49 \;\equiv\; 1^{502}\cdot49 \text{ (mod 100)} \;\equiv\; 49 \text{ (mod 100)}

    The difference is: . |4-9| \:=\:5

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the sum of the digits of the number
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 5th 2012, 05:48 PM
  2. difference of b to base 2010 and b to base 2009
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 19th 2010, 07:16 AM
  3. Replies: 3
    Last Post: August 2nd 2009, 03:00 PM
  4. Finding last two digits of a large number
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 29th 2009, 01:30 AM
  5. Finding digits of a prime
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: October 3rd 2007, 12:48 AM

Search Tags


/mathhelpforum @mathhelpforum