I need some help..

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Tbone
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I need some help..

Post by Tbone »

i'm just stuck and can't stop thinking of this one.. can anyone help me?

One piece of paper is cut into 8 pieces. We select X pieces from them and cut each into 8 pieces. We select again X pieces and cut each into 8 pieces…continue doing like this. So which of the following can be a possible number of pieces in total after cutting?

2003?
2004?
2005?
2006?
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Karura
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Post by Karura »

the answer have to be an even number, at least I think ;)

my math skillz is equal to zero.
Last edited by Karura on 15 Feb 2010, 22:57, edited 1 time in total.
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Tbone
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Post by Tbone »

That is wrong, I just calculated the right answer and it is 2003.
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Gorlom
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Post by Gorlom »

karura 8 - X + X*8 describes what happens after you take the first X number pieces and cut them into 8 pieces. "X*8" will allways be even, but if X is an odd number 8 - X will also be an odd number. Which means 8 - X + even number can be an odd number.

I'm not exactly sure what formula one would use to calculate the continued cutting X pieces of paper up though. But it will be up to the number of times you keep cutting up X pieces wether or not the number is odd or even.

(for examlple if X = 3 you first get 29 and after that 50 and after that 71, 92 etc etc
X=5 you get 43, 78, 113 etc)

hmmm 8-X + X*8 + X*7*(Y-1) where Y is the number of times you take X pieces of paper and cut them into 8.

Does that work sinu or what formula did you use? I'm sure there's a simpler one. But I haven't done this for some time and I kinda just woke up after a nap =P
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Tbone
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Post by Tbone »

On the first round of cuts, you take x_1 number of pieces out of the original 8 and you cut those into 8. So in the first round you have a total of:
8 - x1 + 8*x1
= 8 + (8-1)*x1
= 8 + 7*x1

On the second round of cuts, you take x_2 number of pieces from what you had in the last round and then you cut those into 8 pieces. So then you have a total of:
8 + 7*x1 - x2 + 8*x2
= 8 + 7*x1 + (8-1)*x2
= 8 + 7*x1 + 7*x2

If you extrapolate, for xN number of rounds, you will have
8 + 7*x1 + 7*x2 + 7*x3 + ..... + 7*xN

Which has to equal one of the choices. Or, better yet, the following has to hold true:
8 + 7 * (x1 + x2 + x3 + ..... + xN) = 2003, 2004, 2005 or 2006

Since x1 + ... + xN has to be a whole number, then you're left with the only answer which is 2003.
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Gorlom
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Post by Gorlom »

Does X stay the same each time? I'm confused about how to read that x1, x2, x3 stuff.

ugh I forgot all about that annoying N crap math... or perhaps i never really understood it in the first place.
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Graav
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Post by Graav »

X is a variable so it doesn't stay the same, Tbone's math is correct nice one :p
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