Puzzle with a DifferenceDate: 03/19/2003 at 22:55:37 From: Janice Subject: It's a Puzzler Place each number from 1 through 10 in a box. Each box must contain a number that is the difference of two boxes above it, if there are two above it. The underscore represents the boxes that are set up as follows: _______ _________ __________ __________ _________ ___________ _________ __________ ___________ _____________ 10 3 8 9 7 5 1 2 4 2 Thank you for your help. Janice Date: 03/20/2003 at 22:25:42 From: Doctor Peterson Subject: Re: It's a Puzzler Hi, Janice. This is a sort of puzzle for which there is not a lot of help I can give other than to encourage persistence. Especially at your level, it is probably meant just to give you a lot of practice with subtraction; the bad part is that it can be very frustrating. You need a way to be orderly in your testing. I tried playing with it, and found three solutions without filling more than a page with attempts. Maybe some of my tricks can help you, though I wouldn't expect someone without a lot of math experience to think of all of them. One thing I did was to think about odd and even numbers. When you subtract two numbers, whether the difference is odd or even depends only on whether the numbers you started with are odd or even. So I could see what sort of pattern of odd and even numbers we will get by just using 0 and 1 as my sample odd and even numbers. There are 16 possible patterns of odd and even to start with; here are three of them: 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 1 10 even 6 even 5 even Now, the first two of these won't work for the numbers 1-10, because we have five even and five odd numbers, and these don't. But the third does have the right number of odd and even numbers. So if we start with even-even-odd-even, we have a chance of using all ten numbers. The only other patterns that will work are even-even-odd-odd, even- odd-even-odd, and the reverses of these three (which we can ignore because if we discover a solution that fits one pattern, we can reverse it to make the other). That helps limit what patterns I try. (My three solutions happen to be one in each pattern; I wouldn't have expected that.) Another important fact is that 10 has to be in the top row. Do you see why? There is no pair of numbers that could go above it. And 9 can't be below the second row, and even there it can appear in only one way, as the difference of 10 and 1. So it helps to start at the top row with the large numbers. In two of my solutions 9 was in the top row, along with 8. One more idea is to save writing by filling in just a part of the triangle and then trying all the possibilities for what's left without writing. For example, suppose I try 10, 2, 9 in the top row, leaving a space either on the left or the right of them for the fourth. I get this: 10 2 9 8 7 1 Now the numbers I have left are 3, 4, 5, 6. I can try each of these in each end of the top row to see what will happen, and quit as soon as I get a repeat number. For example, if I try 3 on the left I get 3 10 2 9 7 8 7 . 1 . and already I can quit. If I try 3 on the right, I get 10 2 9 3 8 7 6 1 1 . and I can quit now. If you're systematic enough, you might be able to go through all the possible groups of three in the top row; but most likely you'll have an answer or two before you get too far. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 03/21/2003 at 00:20:21 From: Janice Subject: Thank you (It's a Puzzler) Thank you! Your advice about persistance paid off for my daughter and the Pascal Theory as well. She was so excited when she got the answer of: 8 10 3 9 2 7 6 5 1 4 and of course, that reversed. We still have not gotten the 3rd way you mentioned, but we're still playing with it. Thank you again. Janice Date: 03/21/2003 at 08:38:53 From: Doctor Peterson Subject: Re: Thank you (It's a Puzzler) Hi, Janice. Good job! That's actually the first one I found. Its mirror image was not one of my three; I didn't count that because I consider it the same solution. Another math doctor gave me a list of ALL solutions (probably found by computer); he has four solutions, again ignoring mirror images, so I only missed one. Interestingly (and I'll have to think about the math behind this), the solutions come in pairs, with the same top four numbers in different orders. Two of mine formed a pair, and his fourth solution pairs up with my third solution. So you should be able to find a second easily, and two more with persistence. It's an interesting puzzle after all, isn't it? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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