Number Puzzle with Digits 1-9 in a 3 by 3 GridDate: 10/09/2006 at 19:50:55 From: Mike Subject: Math Puzzle In a 3 cell by 3 cell grid, use the digits 1 through 9 to fill in the cells so each horizontal, vertical, and diagonal row of three has a sum of 15. No digit may be used more than once. I've tried multiple ways with no solution! Frustration! This puzzle is my 4th grader's POW (Parent Over Worked). I have no idea why my son's math teacher would send something like this home other than to have the parents do it. Date: 10/09/2006 at 22:43:28 From: Doctor Peterson Subject: Re: Math Puzzle Hi, Mike. This is a standard (and ancient) puzzle called a Magic Square. I'm not sure what the point is in assigning it, myself: many people know about it and can either look it up or build it using some fairly simple rules; those who don't just have to use lots of trial and error. I don't know how one would work it out using any specifically "mathematical" ideas, apart from those very specific techniques for building magic squares. I would hope, since it is so unfair as a test of any particular skill, that its purpose is simply FUN. I'm sure it's not meant to frustrate kids or make their parents do a lot of work. Enjoyment of puzzles is one of the foundations of later math skill! I'll pretend I've never heard of the puzzle, and try talking through what I might do to solve it by trial and error. Maybe that will give you a start at how to guide your child to work on it, and perhaps even enjoy it. I'll just draw the grid as _ _ _ _ _ _ _ _ _ Let's suppose we just start by trying the 1 in the left corner: 1 _ _ _ _ _ _ _ _ We have to be able to get 15 in the horizontal row, the vertical row, AND the diagonal containing the 1! So we need to make 14 using two numbers, in three different ways. Let's see ... that would be 9+5, 8+6, and 7+7 -- oops! We can't do it, since the numbers have to be different. So 1 can't be in a corner. We've learned something important! How about putting 2 in the corner; then we need three pairs of numbers that each add up to 13 so that the total in each direction is 15: 2 _ _ _ _ _ _ _ _ We can use 9+4, 8+5, and 7+6. That looks better. But where shall we put them? Let's just try one possibility, and see what goes wrong that we could fix: 2 9 4 7 8 _ 6 _ 5 First, I see we've made a diagonal; does that add up to 15? 6+8+4 = 18, not 15. So maybe the first thing we have to do is to pick one number from each pair to go on the diagonal. We need to take away 3 from the sum; we could do that by just swapping the 8 with the 5. And in fact, that's the ONLY way without moving the pairs themselves to new places, since swapping in the other pairs will always INCREASE the sum. So now we have 2 9 4 7 5 _ 6 _ 8 Four of the six sums are correct! I've taken you almost all the way to a solution, though maybe not to the only one, as far as we can tell, since there are a lot of rearrangements I didn't try (putting a different pair than 5,8 on the diagonal, for example, or putting 3 in the corner). I'll let you finish, and then try guiding your child to find a solution. The sort of thinking I've just demonstrated is basic mathematical reasoning--we go through possibilities in an orderly way, looking at what happens and making adjustments. I don't know whether this sort of thing is taught in 5th grade (or, perhaps, can really be taught at all); but I think the main benefit in assigning puzzles like this is the opportunity for students to discuss their reasoning and develop the skill of talking about how they think. This "reflective thought" is an important part of good learning, and I hope your child gets a chance to do it, either with you or in class or with fellow students at other times. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 10/11/2006 at 12:08:07 From: Mike Subject: Re: Math Puzzle Doc, Thanks for the help and the quick response. Sometimes things that seem simple can be very frustrating! With your very detailed instructions it made it understandable. Have a great day, Mike |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/