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Grid Patterns

Date: 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   

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:



Now cancel out the common terms and you get the following:


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   
Associated Topics:
Middle School Algebra

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