164067

At first I looked at the problem and thought there were 10 rectangles (all the different coloured small rectangles). I soon realised that there would be many rectangles inside the larger rectangle and the larger rectangle as well so there would be a lot more than 10.I started by focusing on the first rectangle there was only one. I then went on to the second rectangle. I realised that there would be one rectangle that was the first two squares and a rectangle which was just the second rectangle. I then came up with a theory, my theory is that what ever rectangle you are on the rectangles that fit inside the rectangles before and the rectangle you are on would add up to how may rectangles would fit inside the current one. For example, take the fourth rectangle you would start by adding how many rectangles fit inside the first rectangle. Only one. You would then take the take the second rectangle. 2 new rectangles fit inside the second rectangle. If you add them together and then add them on to 3 since that is how many new rectangles fit inside the third rectangle you get 6. 6 rectangles fit inside the third rectangle. You then add this in to 4 since in the fourth rectangle you get 4 new rectangles. If you add 6 and 4 you get 10. This means that 10 rectangles fit inside the fourth rectangle. This is the formula to solve a problem like this. This would work all the way up to the hundreds and thousands.