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Curly05
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GMAT 1: 750 Q49 V42
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The simplest way to solve this is to draw a 4x4 square in the top-left-hand corner of the 8x8 square. NOw imaging sliding it over one square at a time to the right. ONce you get to the end, imagine sliding it one square at a time down. If you know how to count, you should be able to solve this problem.
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AkamaiBrah
The simplest way to solve this is to draw a 4x4 square in the top-left-hand corner of the 8x8 square. NOw imaging sliding it over one square at a time to the right. ONce you get to the end, imagine sliding it one square at a time down. If you know how to count, you should be able to solve this problem.



Isn't the same as the amswer I posted erlier ? :?
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Not really. 8C4 is a big number and assumes that you can break up the 4x4 square into columns of 1. If you slide the 4x4 square across the top to the right, you can only move 4 times before hitting the other end.
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My guess is 25, please let us know the anwer. :oops:
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My answer is 49. :-D
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AkamaiBrah
Not really. 8C4 is a big number and assumes that you can break up the 4x4 square into columns of 1. If you slide the 4x4 square across the top to the right, you can only move 4 times before hitting the other end.


Based on this method I get 36. Number of 4x4 sqares along the horizontal line is 6. Obviously, it will be agin 6 along the vertical line.

so 6 x 6 = 36. What is wrong with this? Thanks
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AkamaiBrah
Not really. 8C4 is a big number and assumes that you can break up the 4x4 square into columns of 1. If you slide the 4x4 square across the top to the right, you can only move 4 times before hitting the other end.


Akamai,

Why woudld we be able to move 4 times only? Wouldn't we be able to move 6 times? I thought while moving towards the right each time we would consider the second set of TWO small vertical squares part of our next 4x4 configuration. Please clarify. Thanks



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