Thinking Outside the Pen

Date: 09/28/2004 at 19:40:26
From: Dana 
Subject: Critical thinking problem

How can you put 21 pigs in 4 pig pens and still have an odd number of 
pigs in each pen?  We cannot figure out a way to get the answer, 
because adding 4 odd numbers will give you an even number.

Date: 10/01/2004 at 07:55:40
From: Doctor Beryllium
Subject: Re: Critical thinking problem

Hi Dana, 

You are quite right when you point out that adding 4 odd numbers gives 
an even number.  We can prove it as follows:

Suppose we have 4 odd numbers: 2a+1, 2b+1, 2c+1 and 2d+1.  Adding 
these together we get:

    (2a+1) + (2b+1) + (2c+1) + (2d+1)

  = 2(a + b + c + d) + 4 

  = 2(a + b + c + d + 2)

This shows that the sum of four odd numbers is divisible by 2 and 
hence is not odd, no matter what the numbers are. 

However we should give some thought to how the pig pens are 
constructed.  We naturally assume that the pig pens are laid out as 
follows:

    +-------+-------+
    |       |       |
    | pen 1 | pen 2 |
    |       |       |    
    +-------+-------+
    |       |       |
    | pen 3 | pen 4 |
    |       |       |
    +-------+-------+

But this is just an assumption. The pig pens do not _have_ to be laid
out in this way.  For example if we construct the pig pens as follows:


   +---------------+------+
   |   pen 1       |pen 3 |
   |  +------+     |      |
   |  |pen 2 |     +------+
   |  +------+     |pen 4 |
   |               |      |
   +---------------+------+
   
Then one solution to the problem would be,

   pen 2: 1 pig
   pen 3: 7 pigs
   pen 4: 7 pigs
   pen 1: The 6 remaining pigs go in pen 1 but not in pen 2.

Now we could argue that pen 1 contains 7 pigs: 6 in its outer region 
and 1 in the inner region.  Thus pen 1 contains an odd number of pigs.

Using this method of reasoning there are many solutions to the 
problem.  Can you think of another one?

Feel free to write back if you need more help.

- Doctor Beryllium, The Math Forum
  http://mathforum.org/dr.math/