# help needed urgently on maths c/wk

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#1
i really need help on finding the rule in the number grid cwk
0
17 years ago
#2
Number grid coursework? Haven't heard of that one. What's it about?
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17 years ago
#3
If you show me what the number grid is, I can show you how to get the rule.
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#4
its where u get a 10x10 grid and then pick out grids from it( like a 2x2 grid) then u have to multiply the opposite corners and find the differnce of the answers
u have to experiment with squares and rectangles
i,ve done the squares but the rectangles is sooooooooooooo confusing!!!
heeeeeeeeeellllllllllllppppppppppp!!
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17 years ago
#5
Just give me 10 minutes or so to do some working. Even with my PhD I am not a computer!
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#6
(Original post by chrisbphd)
Just give me 10 minutes or so to do some working. Even with my PhD I am not a computer!
oh thanx!!!
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17 years ago
#7
(Original post by chrisbphd)
Just give me 10 minutes or so to do some working. Even with my PhD I am not a computer!
OK. Consider the following 3x2 rectangle:

15.....16.....17
25.....26.....27

Let n=15, this makes the rectangle the following:

n..........n+1.......n+2
n+10.....n+11.....n+12

The difference is therefore:

(n+10)(n+2) - n(n+12)
= [(n^2)+12n+20]-(n^2)-12n
=20

This is true for all 3x2 rectangles in the number grid - it is an algebraic proof.

A similar method can be used for other shapes.

I hope this helps.
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#8
thanx but i need an overall rule, where u can find the difference of any sized grid ( within a 10x10 grid) by using that equation

i think it's 10(L-1)(w-1)....................L=length and w= width

but i dont know how i got it!!!!
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17 years ago
#9
ok 10 more minutes
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#10
(Original post by chrisbphd)
ok 10 more minutes
u'r sooooo kind!!!
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17 years ago
#11
You are spot on with your rule d = 10(L-1)(W-1).

The (W-1) arises from the fact that the difference in numbers along the horizontal of the rectangle is always the width - 1. The 10(L-1) arises from the fact that the difference in numbers along the vertical of the rectangle is always 10 times the length - 1. This is because the numbers increase by 10 each time going down the 10x10 square. Multiplication of these two terns does give the difference you are looking for.

That was a bit waffly but it's difficult when you dont have the student in front of you to explain it! I hope it clears the matter up.
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#12
hey thanx! i understand it now
but wot workin out wud u use to prove wot u have found
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#13
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17 years ago
#14
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17 years ago
#15
(Original post by Saf!)

Could you not just use the agebriac proof from before:

n.............n+1.............. n+2...................... ..........n+(W-1)
n+10........n+10+1......... n+10+2..
n+20........n+20+1
..............
n+10(L-1)..n+10(L-1)+1 ................................... .....n+10(L-1)+(W-1)

Taking these values:

(n+(W-1))(n+10(L-1))-n(n+10(L-1)+(W-1))=n^2+n(W-1)+10n(L-1)+10(W-1)(L-1)-n^2-10n(L-1)-n(w-1)=10(W-1)(L-1)
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17 years ago
#16
I have checked that and it seems fine to me. I had a mental block for a minute! Not a good example to set!

Well done and thanks.
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17 years ago
#17
(Original post by chrisbphd)
I have checked that and it seems fine to me. I had a mental block for a minute! Not a good example to set!

Well done and thanks.
Hey, was something interesting to do. Also Saf!, further points of investigation would be to look at what the effects would be of using a wider or narrower grid (follow the same method and think about the 10(L-1) bit). Also bear in mind that there will be a restraint on W dependent on the width of the grid and the value of n.
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#18
i dont understand how u simplifiied that bit!

(n+(W-1))(n+10(L-1))
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