Date: Mon, 11 Sep 2000 21:14:56 -0500 (CDT)
From: Mark Jacobson <jacobson@cns.uni.edu>
To: 810-080-01@uni.edu, 810-080-02@uni.edu
Subject: Consecutive Integers hints...

> 
> I have been trying to figure out #22 of the homework on page 93.  It 
> is:  Prove that "The sum of 3 consecutive integers is divisible by 3."

  Hints from the instructor in response to email question:
  (See web page if you want to read from your web browser)

    Let x, y and z be the 3 consecutive integers:

    You can say x, y and z where x < y < z and the following is true,
                                               y = x + 1 and 
                                               z = y + 1
    *** Two cases:

    #1.  where x is even,           x = 2m              for some integer m
               y is x + 1           y = x + 1 or 2m + 1
          z is x + 2 or y + 1       z = y + 1 or (2m + 1) + 1
                                      
    #2.  where x is odd, 

          x = 2r + 1 for some integer r
          y = x + 1 = (2r + 1) + 1
                      (xxxxxx) + 1

          z = y + 1 = ((2r + 1) + 1) + 1
                      (yyyyyyyyyyyy) + 1

     In either case, you just have ONE variable to deal with now.
     The variable m or r or w or whatever you called the integer.

     The rest is just seeing what the sum (x + y + z) is when you 
     put all of the m's or r's together as you re-express the sum x+y+z.

     In both cases, it should not be impossible to show how a
     3 can be factored out of the mess!