154691

I figured out the problem by first finding the area of all of the squares. Once I figured out the area of the squares, I added all of the sums together. At first I got 956 for the area, but when Mr. Owens showed us what other students had done, the right answer for the area was shown on the screen. It turned out that I had forgotten to add in the area of the 10 by 10 square (100). I then figured out that you have to find all of the factors for the total area. Mr. Owens showed us all of the factors for 1056, which is the total area. He also mentioned that the sides of the rectangle could be bigger than 18 (which is the dimensions of the biggest square). I tried dividing that by 18, but the answer turned out to be 58 r12. After I had listed them down, I thought that 32 x 33 would be the most likely of all. I then drew a rectangle with those dimensions, and tried experimenting. At first, I wasn't sure where to put the 18 square, but my friends Dakota and Hannah suggested that I put it in a corner. So I started by putting the 18 square in the top left corner. Since 18 + 18 is 36, I knew that 18 + 14 is 32. Therefore, underneath the 18 square, I put the 14 square. I put the 4 square right underneath the 18 square on the right side. Then, since 4 + 10 is 14, I added the 10 square right underneath it. I realized that the nine square could fit beside the 10 square, so I put it there. I made a mistake at first and made the dimensions of the 9 square 9 by 10. Since it looked equal, I thought I was on to something. After that, I tried to put the 15 square on top of the 10 square, but I knew it wouldn't work. So then I tried the seven square. It fit well, and there was 15 squares left on top of it. There were 8 squares left over. It was perfect for the 8. But, I realized that the 8 was a square taller than the 7. That was when I figured out that I had made a mistake. So I corrected the 9 and 8 squares, and found the perfect spots for the 15 and 1 squares. Then, I realized that it all fit!