Results 1 to 8 of 8

Math Help - Dividing Integers

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    8

    Dividing Integers

    I have run across some interesting properties of the quotient obtained from dividing 2 integers. I will list the properties in which I am interested at the conclusion of this post. The question is: Where can I find proofs of these properties.

    Here are the properties:

    1. An integer is divisible by 3 if the sum of its digits is divisible by 3.

    2. An integer is divisible by 4 if the last 2 digits of the number form a number which is divisible by 4.

    3. An integer is divisible by 8 if the last 3 digits of the number form a number which is divisible by 8.

    4. An integer is divisible by 9 if the sum of its digits is divisible by 9.

    Thanks for any input.

    ... doug
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Dividing Integers

    The proof of 1), for a three-digit number "abc"
    abc = 100a + 10b + 1c
    = 99a + 9b + (a + b + c)
    3 divides 99a + 9b, so 3 divides "abc" iff 3 divides (a + b + c), the sum of the digits.
    You can do this for numbers of any length, but I'm not latexing that from my iPhone.

    2). Express any number as "abcdef00" + "gh" (for example).
    Since "abcdef00" is a multiple of 100, it is divisible by 4. So all that remains is to check "gh", the last two digits.
    3) same as 2), since 8 divides 1000
    4) same as 1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2011
    Posts
    8

    Re: Dividing Integers

    Awesome! Thanks.

    ... doug
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Dividing Integers

    You can work out other divisibility tests for, say, prime numbers (the building blocks on the integers, if you will).
    It gets complicated quickly, though!
    Consider divisibility by 7. We look for multiples of 7 that are CLOSE TO POWERS OF TEN (ten, since we use base ten).
    It gets nasty quickly, but stick with me!
    1 + 0 = 1
    3 + 7 = 10
    2 + 98 = 100
    (-1) + 1001 = 1000
    (-3) + 10003 = 10000
    (-2) + 100002 = 100000
    The second number in each equation is divisible by 7. So for a six digit number "abcdef", we write this as:
    a*100000 + b*10000 + c*1000 + d*100 + e*10 + f*1.
    =
    (a*100002 + b*10003 + c*1001 + d*98 + e*7) + [-2*a - 3*b - 1*c + 2*d + 3*e + 1*f]
    Now everything in (...) is divisible by 7. It remains to check that the contents of [...] are divisible by 7. It's not as clean as just adding up the digits, but you have a method.
    p.s. the pattern of coefficients in the brackets [-2,-3,-1,2,3,1] repeats so you can extend this to arbitrarily large numbers.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Dividing Integers

    Now, see if you can work out something similar for divisibility by 11. It's fairly simple, since
    1 + 99 = 100
    (-1) + 1001 = 1000
    1 + 9999 = 10000
    (-1) + 100001 = 100000, and so on.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2011
    Posts
    15

    Re: Dividing Integers

    To summarize (2) (3): 2^n divides a number if the last n digits can be divided by 2^n.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Dividing Integers

    Where can I find proofs of these properties?
    /summary
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2011
    Posts
    15

    Re: Dividing Integers

    For number 4: I don't have time to prove it at the moment, but you can recursively add the digits of a number divisible by 9 to get the number 9. That might be a hint for the proof. I'll work on it too.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 5th 2011, 11:47 PM
  2. Replies: 7
    Last Post: August 3rd 2010, 01:31 PM
  3. Matrix of integers whose inverse is full of integers
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 19th 2010, 02:02 PM
  4. Replies: 4
    Last Post: February 24th 2008, 03:08 PM
  5. Replies: 2
    Last Post: October 14th 2007, 03:18 PM

Search Tags


/mathhelpforum @mathhelpforum