Results 1 to 3 of 3

Math Help - modular arithmetic

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    10

    modular arithmetic

    Determine the last digit (the ones digit) in the base 8 expansion of 9^1000, 10^1000, and 11^1000.

    I'm so confused is there a trick to doing this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by demon1 View Post
    Determine the last digit (the ones digit) in the base 8 expansion of 9^1000, 10^1000, and 11^1000.

    I'm so confused is there a trick to doing this problem?
    9 = 8+1

    so look at (8+1)^1000, all the terms in the binomial expansion of this are
    multiples of powers of 8 except for the last one, and so do not contribute to
    the least significant digit base 8 of the result. So only 1^1000 contributes to
    the ;east significant digit, but this is 1, so the least significant digit is 1.

    11= 8+3

    For the same reasons as above the least significant digit of 11^1000 is 3^1000
    in base 8. But 3^1000 = 9^500, so in base 8 3^1000 = 9^500 which has least
    significant digit 1 for the same reason as in the first part.

    10=8+2

    So using the same sort of argument as in the first part we know that the least
    lsignificant digit base 8 is the least significant digit base 8 of 2^1000 base 8.

    2^1000 = 2.2^999 = 2.8^333

    so the least significant digit of 2^1000 in base 8 is 0.

    RonL

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,907
    Thanks
    766
    Hello, demon1!

    Determine the last digit (the ones digit) in the base 8 expansion of:
    . . (a) 9^1000 . . (b) 10^1000 . . (c) 11^1000

    (a) Write nine in base-eight: .9 = 11
    8

    If a number ends in 1, powers of the number will also end in 1.
    . . 11 = 121
    8, .11 = 13318, .11^4 = 146418, . . .

    Therefore: 9^1000 =(11
    8)^1000 ends in 1.


    (b) Write ten in base-eight: .10 = 12
    8

    Then: .(12
    8) = 1448, .(128) = 17508

    Thereafter, all powers of 12 will end in 0.

    Therefore: 10^1000 = (12
    8)^1000 ends in 0.


    (c) Write eleven in base-eight: .11 = 13
    8

    Then: .13 = 13, .13 = 171
    8, .13 = 24638, .13^4 = 344618, . . .

    The ones-digits alternate: 3, 1, 3, 1, . . .

    Therefore: 11^1000 = (13
    8)^1000 ends in 1.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Modular Arithmetic
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 28th 2010, 06:08 AM
  2. Arithmetic Modular help!!
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 24th 2010, 09:44 AM
  3. modular arithmetic
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 2nd 2009, 01:17 PM
  4. Modular arithmetic
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: December 13th 2008, 04:17 PM
  5. Modular arithmetic help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 4th 2008, 09:47 AM

Search Tags


/mathhelpforum @mathhelpforum