Number PuzzleDate: 02/12/97 at 11:22:02 From: marko kuo Subject: problem solving I was given a number puzzle to solve in my problem solving class and I kind of got stuck. I suspect there might be a theory behind it. The problem is to find a number with 9 digits in which: the first digit is the number of 0's in this number the second digit is the number of 1's in this number . . the 9th digit is the number of 8's in this number Thanks, Kuo Date: 02/13/97 at 15:39:35 From: Doctor Wilkinson Subject: Re: problem solving This is really quite tricky. Don't feel bad about getting stuck. I got stuck a couple of times myself. What I did was write a heading for each digit: 0 1 2 3 4 5 6 7 8 I then started filling in possibilities to try to narrow things down. Suppose we had: 0 1 2 3 4 5 6 7 8 2 Then we need to have an 8 in two places, but that gives us 8 of two different digits, and we only have 9 digits. So under the 8 we must have a 0 or a 1, and similarly for 5, 6, and 7. Now let's try: 0 1 2 3 4 5 6 7 8 1 Now we've got to put that 8 somewhere, which means that there's 8 of something, and there just isn't room, because there are only 7 places left. So we must have: 0 1 2 3 4 5 6 7 8 0 Moving on to the 7, suppose we had: 0 1 2 3 4 5 6 7 8 1 0 Then that 7 has to go someplace, and we also have a 1 already, and again there isn't room. So we have: 0 1 2 3 4 5 6 7 8 0 0 And let's suppose we have a 1 under the 6. This gives us: 0 1 2 3 4 5 6 7 8 1 0 0 Now there isn't room for 6 of anything except 0, so we have to have: 0 1 2 3 4 5 6 7 8 6 1 0 0 We then have to put the 1 someplace. It can't go under the 1, because that would give us two 1's. But if it doesn't go under the 1, then we have to have a 2, 3, 4, or 5, and once again we don't have room. So we have: 0 1 2 3 4 5 6 7 8 0 0 0 Let's try: 0 1 2 3 4 5 6 7 8 1 0 0 0 Again, you have to put the 5 under the 0: 0 1 2 3 4 5 6 7 8 5 1 0 0 0 Now you can't put the 1 under the 1, just like last time. So let's put a 2 under the 1 and see what happens: 0 1 2 3 4 5 6 7 8 5 2 1 0 0 0 Now we have a 2, so we need to put something other than 0 under the 2, which leaves us no choice but to put the two more 0's we need under the 3 and the 4, and there's only room for one 2, which gives us: 0 1 2 3 4 5 6 7 8 5 2 1 0 0 1 0 0 0 It works! You should finish this by eliminating any other possibilities. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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