Thoughtful Trial and ErrorDate: 05/21/2002 at 23:42:17 From: Ashley Subject: It will make you dizzy! Dear Dr. Math, Here is the problem I need to figure out: O O O O O O O O O O Place the numbers 1 through 10 in the circles so that: a) the rows of three circles have the same sum; and b) the rows of four circles have the same sum (not necessarily the same sum as the rows of three circles). I have tried trial and error for hours on this annoying problem. Is there another way to figure it out? Please help! Thanks, Ashley Date: 05/22/2002 at 23:06:15 From: Doctor Peterson Subject: Re: It will make you dizzy! Hi, Ashley. To tell the truth, this kind of puzzle is largely trial and error, so it's hard to give hints without giving the answer. But I just sat down and solved it in a few minutes, using what you might call experienced trial and error. I'd like to show you some of what I did. One trick I know for puzzles like this is to think about what happens if you add up all the rows together. In this case, if you add the three rows of four (whose sums I'll call A) and the three rows of three (whose sums are B), you have added in each number twice, except that the middle one is counted three times. But the sum of the numbers from 1 to 10 is 55. So 3A + 3B = 2*55 + X where X is the middle number. That limits what X can be: since 2*55 = 110 is one less than a multiple of 3, X must be one more than a multiple of 3: 1, 4, 7, or 10. So now let's choose a middle number; I picked 10 to try first. (Since there's no number right in the middle between 1 and 10, an extreme seemed like a good choice!) Then 3(A + B) = 110 + 10 A + B = 40 I just chose a neat set of numbers for the three corners; taking the four used numbers out of the set, the sum of the rest of the numbers is 55 minus their sum. A third of that sum tells me what the sum of opposite numbers has to be; and I could tell from the sums of the pairs of numbers at the corners, what sums I needed for the middle two numbers along each side (that is, three different sums). Then I just tried placing the six remaining numbers in their places (with the corner numbers _not_ yet written) to fit those requirments, and then I could tell where the corner numbers had to be. My first choice turned out to work; that suggests that there may well be other solutions. I know there will be a solution with 1 in the middle (obtained from my solution by subtracting each number from 11). There may be still more. Have fun trying for your own solution! I hope this gives you some good ideas, without taking away the fun of discovery, and a lesson about thoughtful trial and error. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/23/2002 at 08:53:13 From: Doctor Peterson Subject: Re: It will make you dizzy! Hi again, Ashley. When I picked up my papers last night I looked at my solution and realized it was wrong. In case that misleads you, I want to let you know that I did find a solution, but it used 7 in the middle rather than 10; I think I'm convinced that 10 in the middle doesn't work. Also, when I found the real solution I used a different trick than I had mentioned. Since I had determined the sum A+B for a given middle number X, I realized that I could subtract that sum from 55 and find the sum of the three numbers at each corner. Then I could find all the ways to divide the remaining numbers into three groups with that sum (there were only two ways!), and then try each of the three possible placements of one of those groups and see whether it led to a solution. That's how I convinced myself there is no solution with X=10. The puzzle is harder than I thought, but with careful trial and error it can be done! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/25/2002 at 19:52:17 From: Ashley Subject: It will make you dizzy! Question submitted via WWW: Hey Dr. Peterson, Thanks so much. I finally figured it out! Yay! Thanks again, Ashley |
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