Grid PatternsDate: 09/10/99 at 11:04:16 From: Sunny Patel Subject: Math and Algebra At school I have been assigned an investigation. It is called "grid patterns." What you do is the following: start off with the grid: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 draw a 2*2 square around any 4 digits, for example: +------+ | 1 2 | | 6 7 | +------+ cross-multiply the digits: 1*7 = 7 2*6 = 12 now take the difference: 12-7 = 5 If you try the same with another 2*2 group of numbers the difference is still 5. We have to work out a general formula that will determine the difference for all squares and rectangles, and then we have to write the formula in algebraic terms. Please help me! Date: 09/10/99 at 12:07:17 From: Doctor Peterson Subject: Re: Math and Algebra If your grid always has five numbers per row, then the number just below any number is 5 more. If we make a rectangle A numbers wide and B numbers high, it will look like this: ... x x+1 ... x+A-1 ... x+5 x+6 ... x+A+4 : : : x+5(B-1) x+5(B-1)+1 ... ??? (If this isn't clear, try setting A and B to 2 and you should get the correct numbers for the 2 by 2 squares in your example.) Figure out what that lower right corner is, then cross-multiply and see what you get. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/13/1999 at 11:57:09 From: Sunny Patel Subject: Math and Algebra Dear Dr. Math: I wrote to you that I had a problem with an investigation. Do you remember? You helped me work out the problem. Next, I worked with squares with rows of 6 and the difference was 6, and with rows of 7 you get a difference of 7. My teacher said that I should work out a formula that would work out the difference for all squares and rectangles. So over the weekend I wracked my brains trying to think of a formula that would work. A friend showed me the following formula for rectangles. You have a rectangle and in the 4 corners you have this: n n+(x-1) : : n+g(y-1) n+g(y-1)+(x-1) Next cross-multiply the above and you get the following: n^2+ng(y-1)+n(x-1) n^2+n(x-1)+ng(y-1)+g(y-1)(x-1) Now cancel out the common terms and you get the following: g(y-1)(x-1) and that is the answer. I do not understand how you got this, so in my report, how would I explain how to get this? Please help me! Date: 09/13/1999 at 12:57:58 From: Doctor Peterson Subject: Re: Math and Algebra Hi, Sunny. What your friend showed you is a lot like what I suggested: >If we make a rectangle A numbers wide and B numbers high, >it will look like this: > > ... x x+1 ... x+A-1 ... > x+5 x+6 ... x+A+4 > : : : > x+5(B-1) x+5(B-1)+1 ... ??? I used x for the number in the upper left, A and B for the width and height of the rectangle, and 5 for the length of each row in the table. Your friend used n, x, y, and g respectively. I'm not sure what part of your friend's solution you are unsure of; I'll try to explain further where the expressions in the rectangle come from. Here's your friend's version: x columns -------------------------------- n ... n+(x-1) | : : | y rows n+g(y-1) ... n+g(y-1)+(x-1) | Here's a specific example, with n = 6, g = 10, x = 3 and y = 4: 3 cols -------- 6 7 8 | 16 17 18 | 4 rows 26 27 28 | 36 37 38 | In the top row, you add one each time you go one space to the right; going from upper left to upper right we have moved x-1 places (2 in my example, with x = 3), so the value is n+(x-1); in my example, 6+(3-1) = 8. In the left column, you add g each time you go down one row, so the bottom left is (y-1) g's more than the upper left, making it n+(y-1)g. In my example, this is 6+(4-1)*10 = 6+30 = 36. Going across the bottom you again add one each place, so the bottom right is again (x-1) more than the lower left, making it n+(y-1)g + (x-1). You can check that this is also (y-1)g more than the top right. The cross-multiplication will be [n+x-1][n+(y-1)g] - [n][n+(y-1)g+(x-1)] and you've done the rest of the steps. If I've missed the point where you have trouble, write back again. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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