## Monday, March 3, 2008

### Math Problem of The Week

Alright, so as you may or may not know, I tutor kids in math on monday afternoons. (this week, on tuesday, and thursday afternoons as well....oy). The particular boys I tutor today always have a "Problem of the Week" which is almost always some very simple or very tedious math problem. They are interesting though, and I usually try and find complex algebraic ways to solve them as opposed to "just countin'" (suck on that Keep-it-simple!). So todays problem looked like this:

How many squares are there?

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:

brad said...

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!