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were to answer is: How many squares are there on an 8 × 8 checkerboard? ... count the squares in a 1 × 1, 2 × 2, 3 × 3 and a 4 × 4 checkerboard.
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D R. D R E W ' S M A T H 2 : P O W W R I T E - U P s
In this POW, we are asked to investigate an 8 × 8 checkerboard and count the number of squares. As shown in the figure on the left below, it is obvious that there are 64 gray and white squares (32 gray and 32 white), but these are not they only squares that can be found on the board. In the problem statement, we were given an example of one 3 × 3 square that can also be found on an 8 × 8 checkerboard, shown in the figure on the right below. The question were to answer is: How many squares are there on an 8 × 8 checkerboard?
When I first started working on this problem, I knew that I would need to use the Stay Organized Habit of a Mathematician; otherwise, it would be really easy to miss a square or possibly even count a square twice. I needed a systematic way of counting ( Be Systematic ) but I really didn’t want to go to the trouble of doing all that counting—it seemed as if it might be a waste of time. Instead, I decided to use the Start Small Habit of a Mathematician and count the squares in a 1 × 1 , 2 × 2 , 3 × 3 and a 4 × 4 checkerboard. It was my hope that a pattern would emerge that would allow me to determine the number of squares in an 8 × 8 (and justify the reasoning, of course). For reference, here are my checkerboards: Starting with a 1 × 1 checkerboard, the number of squares was trivial to count: there is one (and only one) square. For a 2 × 2 checkerboard, there are 2 × 2 = 4 “ 1 × 1 squares” and 1 “ 2 × 2 square” (for a total of three). For a 3 × 3 checkerboard, there are 3 × 3 = 9 “ 1 × 1 squares”, 4 “ 2 × 2 squares” and 1 “ 3 × 3 square” (for a total of 14). To count the number of 2 × 2 squares, I visualized a 2 × 2 square that starts in the upper left and then moves one square at a time to the right. When it hits, the end, it moves back to the left and one square down and repeats. This movement was my way of being systematic and ensuring I was counting correctly. The diagram below illustrates: This approach made it straightforward to count the number of squares in a 4 × 4 checkerboard. There are:
When adding up these numbers (to get a total of 30 squares) I noticed two patterns emerge! First, the number of squares in a 2 × 2 squares is 22 = 4 more than in a 1 × 1 ; the number of squares in a 3 × 3 square is 32 = 9 more than in a 2 × 2 square; and the number of squares in a 4 × 4 square is 42 = 16 more than in a 3 × 3 square. Second, I
When adding up these numbers (to get a total of 30 squares) I noticed two patterns emerge! First, the number of squares in a 2 × 2 squares is 22 = 4 more than in a 1 × 1 ; the number of squares in a 3 × 3 square is 32 = 9 more than in a 2 × 2 square; and the number of squares in a 4 × 4 square is 42 = 16 more than in a 3 × 3 square. Second, I noticed that all the numbers I was adding up were perfect squares. I thought about this for a few minutes and then realized that had to be the case and the method I used to count the number of squares explains why: the number of horizontal and vertical movements of a small square in a larger square has to be the same, so the total number of small squares has to be the square of the total possible number of movements the smaller square can make (vertically or horizontally). For the first pattern, though, it was harder for me to convince myself that the pattern would always work. If I could convince myself ( Seek Why and Prove ), I would know for sure that the answer to the question would have to be: 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204 To help me better understand the pattern I was seeing, I decided to create a table that summarized all the data I had obtained so far: Square Size: 1! 1 2! 2 3! 3 4! 4 5! 5 No. of 1 ( 1 's: 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 No. of 2 ( 2 's: 0 12 = 1 22 = 4 32 = 9 42 = 16 No. of 3 ( 3 's: 0 0 12 = 1 22 = 4 32 = 9 No. of 4 ( 4 's: 0 0 0 12 = 1 22 = 4 No. of 5 ( 5 's: 0 0 0 0 12 = 1 Total: 1 5 14 30 55 Looking at this table, I knew that my first pattern of could not be a mere coincidence; I also knew intuitively that it had to work because of the method I used to count squares (but that in itself is not proof!). After thinking about it for awhile I did realize that my counting method might actual form a proof. Thinking about a really large square, say 100 × 100 , we know that there will be 1002 “ 1 × 1 squares”. We also know that there will be 992 “ 2 × 2 squares” because a 2 × 2 squares is limited to 99 movements horizontally and 99 movements vertically. This pattern we know repeats (because each time we make the square one size larger, it’s horizontal and vertical movements must be one less). Therefore, the 100 × 100 squares must have 1002 more squares than a 99 × 99 square. I became convinced!
By starting small and looking for patterns, I was able to prove that the number of squares in an n ( n checkerboard is S = 12 + 22 + 32 + 42 + ) + ( n + 1 )^2 + n^2 So, for an 8 × 8 checkerboard, the number of squares is S = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204 I was really curious about the expression I derived: S = 12 + 22 + 32 + 42 + ) + ( n + 1 )^2 + n^2 because I knew from work I did last year that this can be written as (using summation notation): S = 12 + 22 + 32 + 42 + ) + ( n + 1 )^2 + n^2 = (^) i = 1 n i^2 This made me wonder if there were a way of calculating the sum without having to go through the trouble of adding up a whole bunch of terms. I gave up really quickly and instead did an online search and came up with: