Numbers on a Star Brain TeaserDate: 01/03/2008 at 22:16:17 From: Linda Subject: 5th grade brain teaser - Mom, Dad & older sister can't solve We have a number sequence of 20, 21, 22, 23, 24, 25, 27, 28, 29 and 31. The numbers are to be placed on a star so that each line of four numbers has a sum of 100. Each number can only be used once. The instructions advise my 5th grade daughter to "guess and check". My husband, 14-year old daughter and me (mom) have tried the guess and check method for about an hour, trying all sorts of combinations. Obviously we all are missing something that is needed to solve the problem. We've tried listing all number sequences that add up to 100. Then we assign numbers to the outside points of the star and use a Sudoku-type method of figuring out what the numbers would be on the inside points of the star. But we're all scratching our heads on this, this is going to be very embarrassing if this is easy. Please help, thank you so much. Date: 01/04/2008 at 10:39:52 From: Doctor Rick Subject: Re: 5th grade brain teaser - Mom, Dad & older sister can't solve Hi, Linda. I don't have any good suggestions beyond what you have done. Well, one minor thing: you can reduce the size of the numbers you have to work with, by subtracting 20 from each number. Then the row-sums are reduced by 4*20 = 80, so we're looking for ways to put the numbers 0, 1, 2, 3, 4, 5, 7, 8, 9, and 11 into the star so the row-sums are 20. I too listed all the sets of 4 of these numbers whose sum is 20; I found 13 sets (ignoring order). But it looks like too much work. My instinct, when I can't do much better than guess-and-check and there are a lot of choices to check, is to write a computer program. I did that and found 120 solutions. However, for each distinct solution, there are five ways to rotate the star and another five ways with the star flipped around, so the solutions come in groups of 10. There are 12 distinct solutions, when rotating or flipping the star is not considered to produce a different solution. Note, though, that there are 10! = 3,628,800 ways to arrange the 10 numbers. Taking out the factor of 10 for rotations and flips, there are still 362,880 different stars, of which only 12 are solutions to the problem. I don't know how anyone can be expected to find one of them by guess-and-check, without a lot of luck! Here is the first solution my program found: 24 20 29 28 23 22 21 31 25 27 - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/04/2008 at 21:34:24 From: Doctor Peterson Subject: Re: 5th grade brain teaser - Mom, Dad & older sister can't solve Hi, Linda. I was going to write an answer almost exactly like the first part of Dr. Rick's answer last night, just suggesting you reduce the size of the numbers and telling you that these problems are inherently hard, so not to worry about it. (I've struggled for a long time with some very similar ones.) But I left it in case anyone could give a more positive answer than that. Then I continued to think about the problem as I brushed my teeth, and within a few minutes I had a solution with remarkably little work. I don't know that a child could be expected to do this (since it's a lot more than mere guess-and-check), but here's an outline of my method: 1. Subtract 20 from each number, and thus subtract 80 from the expected sum of each row; this can be verified by noting that the sum of all 5 rows will be twice the sum of the 10 numbers, so the sum of one row is 2*50/5 = 20. Now our goal is to place 0, 1, 2, 3, 4, 5, 7, 8, 9, and 11 in the star so that each row sums to 20. 2. Place the largest number, 11, somewhere; I put it at the top. That number is in two rows; the sum of the remaining 3 numbers in each row has to be 20-11 = 9. List the ways to make 9 from three of the numbers. 11 / \ / \ x------a-----d------y \ / \ / b e / \ / \ / z \ / / \ \ c f 3. Pair up those five sets of three to find two that do not overlap, since they can't share a number. This gives only two such pairs, namely 0+1+8=2+3+4 and 0+2+7=1+3+5. 4. For each of these, do the following: Consider that the remaining three rows each contain one number from each list (a+d, b+f, and c+e as shown above), combined with two of the remaining three numbers (shown as xyz above). List the three sums formed by pairs of the latter numbers, and from those find the sums that the pairs of numbers from the two sets of three have to add to. For the second pair of sets, the remaining numbers are 4, 8, and 9, giving sums 12, 13, and 17, so that pairs from the two sets have to add to 8, 7, and 3. (The first pair of sets of three don't work; you can't do this step.) 5. Find pairs that give these three sums; there is exactly one way to do this: 0 + 3 = 3 (will be in row with 8+9) 2 + 5 = 7 (will be in row with 4+9) 7 + 1 = 8 (will be in row with 4+8) --- --- 9 9 6. Now place 0, 2, 7 on one row emanating from the 11 (my abc above), and 3, 5, 1 on the other (my def), making sure that the correct pairs lie on rows crossing them. 7. Finally, place 4, 8, and 9 in the spots (xyz) where they give the right sums. 8. Now add 20 to the numbers, and you're done! I think all the solutions are equivalent in the sense that the five rows always contain the same five sets of numbers. Again, don't let anyone feel bad at not getting the answer by guess-and-check; brain teasers are fun when you solve them, but even the best of us don't always do so. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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