So there are more than 64, and at first I just thought it would be multiples of 8 i.e. 1, 2, 4, 8. But then I looked at it and realized there could be 3x3 squares. or 5x5.

This was harder than I thought.

So here comes the whole point, the way I ended up thinking was that basically there are 8 types of squares. 1x1. 2x2. 3x3. 4x4. 5x5. 6x6. 7x7. 8x8. And I know there are 64 1x1 squares and there can only be 1 8x8. "That's funny 64 is 8² and 1 is 1², it's like the opposite."

*cartoonish lightbulb* "Maybe it's the sum of squares up to 8?"

1x1=8²=64 squares + 2x2=7²=49 squares + 3x3=6²=36 squares + 4x4=5²=25 squares + 5x5=4²=16 squares + 6x6=3²=9 squares + 7x7=2²=4 squares + 8x8=1²=1 squares

___________________________________

204 Squares

That's what I figured, and it sounded good to me, can anyone confirm if that is right? Else I'm gonna have to go back over there and pretend that's what *he* wanted to do and that *I* wanted to just count them.

## 1 comment:

It's 204 indeed. I'm not sure what your reasoning behind the solution was, but you could have observed that this is a familiar "embedding" problem, and so starting with a 1x1 square, you have 1 square, with a 2x2 square, you have 4+1 squares, with a 3x3 square you have 9+4+1 squares, with a 4x4 square you have 16+9+4+1, and so on...

This is an example of a series, namely ∑n^2, the sum of which is given (up to finite index n) by: [(n)(n+1)(2n+1)]/6 -- you can check to see indeed that when n=8, ∑n^2=[(8)(9)(17)]/6=1224/6=204.

This gives a more general setting in which you could answer the question when, say, you're given a 41x41 checkerboard.

Hope this helps, and cheers!

Post a Comment