Almost a Magic SquareDate: 12/16/96 at 22:38:59 From: Mike McCabe Subject: 3rd grader stumps the family Hi, My 8-year-old son came home with a problem that we're having trouble with. Is there something we're missing here or is this over the head of most 3rd graders? You have 4 columns and 4 rows. List he following numbers so each row and column totals 44: 3,5,10,11,15,20,8,11,9,10,10,12,13,14,15,10 Is there any easy formula to arrive at the answer? Try and try again is making me look bad. Thank You, Mike McCabe Date: 12/17/96 at 11:03:08 From: Doctor Rob Subject: Re: 3rd grader stumps the family If you can create this arrangement, then you can get another one by rearranging the rows, or by rearranging the columns. Also, you can get another one by making every row a column and vice versa: a b c d a e i m e f g h ---> b f j n i j k l c g k o m n o p d h l p Start with the biggest number, 20. It must be in some row and column. According to the previous paragraph, you might as well assume that it appears in row 1, column 1. Think of what sums can be made from 20 and the other numbers to add up to 44: 44 - 20 - 3 - 5 = 16 impossible 44 - 20 - 3 - 8 = 13 44 - 20 - 3 - 9 = 12 44 - 20 - 3 - 10 = 11 44 - 20 - 5 - 8 = 11 44 - 20 - 5 - 9 = 10 44 - 20 - 8 - 8 = 8 impossible Now the row and column containing 20 must have, besides the 20, six different numbers from the list. Since there is only one 3, one 5, one 8, and one 9, only the following four possibilities are left: 1) 44 = 20 + 3 + 8 + 13 = 20 + 5 + 9 + 10 2) 44 = 20 + 3 + 9 + 12 = 20 + 5 + 8 + 11 3) 44 = 20 + 3 + 10 + 11 = 20 + 5 + 8 + 11 4) 44 = 20 + 3 + 10 + 11 = 20 + 5 + 9 + 10 For each of these, we might as well assume that the first sum containing the 3 lies in the first row, and the second sum in the first column. Since order of rows and columns is not important, we might as well put them in descending order: 1) 20 13 8 3 2) 20 12 9 3 3) 20 11 10 3 4) 20 11 10 3 10 * * * 11 * * * 11 * * * 10 * * * 9 * * * 8 * * * 8 * * * 9 * * * 5 * * * 5 * * * 5 * * * 5 * * * Numbers left: 10,10,10,11,11, 10,10,10,10,11, 9,10,10,10,12, 8,10,10,11,12, 12,14,15,15 13,14,15,15 13,14,15,15 13,14,15,15 Now in each case the problem is reduced to a smaller one. For example, in Case 1, we need to construct a 3-by-3 square of numbers made up from 10,10,10,11,11,12,14,15,15 with row sums 34,35,39 and column sums 31,36,41. Once again, placing the largest numbers, 15 and 15, first, we have only a few possibilities. Continuing in this way, the solution can be found. Mostly it is an exercise in addition and subtraction, together with logic. Once you have found one, using the techniques in paragraph 1, there will be 1151 others (since 1152 = (4*3*2*1)*(4*3*2*1)*2, accounting for the order of the rows, the order of the columns, and the swapping of rows and columns). It can be done! I hope this helps. If you are still having problems, write back. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/