Thinking Outside the PenDate: 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/ |
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