For the first 2 weeks of Math, we explored different methods of growth in Math through the lenses of 4 different problems which were designed not to be hard to solve necessarily, but to provide multiple interesting ways of getting to a correct answer. Additionally, we watched videos from Stanford which focused on the behaviors of the brain when doing math. Each of these videos had a message but the two that stuck out to me was the one which talked about how making mistakes is valuable to growth and the one that said that speed doesn't matter while solving math problems. For much of my academic career, I've been a very fast student so when I'd finish a problem after someone else I'd immediately feel like I was worse than them, same when I'd make a mistake. However, the videos made it very clear that mistakes are nothing but opportunities for growth and being slow means nothing, because it allows for depth. Going slower is something that I've always tried to work on, because going deeper is more of a challenge for me. The first problem we did was Tiling An 11x13 Rectangle (first picture), where we had to cover an 11x13 rectangle with the least possible amount of squares possible. The second was Squares to Stairs (see extension) in which we had to figure out the pattern in squares which would increase in a stair formation each time. The third was the Hailstone Sequence (third picture) which was sequence of numbers which operated by two rules: If the number is even, it gets divided by 2 and if the number is odd it gets multiplied by 3 and 1 is added to it and we had to observe the pattern and deduce whatever we could. The last was the Painted Cubes (second picture) in which we imagined a 3x3x3 cube (made of smaller 1x1x1 cubes) being dunked in paint and then figured out how many cubes had 3 sides painted, how many had 2 sides painted, how many had 1 side painted, and how many had no sides painted. I feel like I did quite well with all the problems and learnt not only how to solve them, but how to approach all mat problems from now on. All in all, these problems were all a lot of fun and really helped me gear up for a year of growth in math!
Work for Tiling an 11x13 Rectangle
Work for Painted Cube
Work for the Hailstone Sequence
Squares to Stairs (Extension)
For this problem we were given three figures (see picture 4 below) and three questions. The first one asked for me to calculate how many squares there would be in figure ten, which was fairly easy. I just drew out the 10th figure and counted the squares which came out to 55. The second question, however, asked for how many squares would be in the 55th figure which was more difficult. At this point, my partner and I had figured out that to find the number of squares in figure 10, you could add 10+9+8+7+6+5+4+3+2+1, and that this method would work with any other figure, so Dr. Drew gave us the task of making it into an equation. To do this I looked at the adding as pairing numbers off to make the figure number plus one, i.e. (in the case of figure 10) pairing off 9 and 2 or 8 and 3 and so on. This ended up forming 5 pairs of eleven which is, obviously, half of the figure number so we make the equation s(number of squares)= (n+1)*(n/2). This then allowed us to answer the last question which asked if there are any staircase structures with 190 squares. This we did using a Habit of a Mathematician: Being Systematic. We used the process of elimination to find out that figure 19 had 190 squares and was a staircase. Since we already knew figure 10 had 55 squares, we were able to make the educated guess that figure 20 would have around 190 squares. It didn't, but it was just barely above which led us to plugging in figure 19 which did have 190 squares. After figuring out the first three questions, I extended the problem further by creating my own pattern which you can see below. It goes from a 1x1 square in figure 1, to a 3x3 in figure 2, a 5x5 in figure 3 and so on. The key thing I noticed was that the dimensions of the square (n*n) were always equal to the figure number plus the figure number minus one, or n+(n-1). From there, I simply adapted that equation so anyone could plug in the figure number (n) and get out the number of squares which made it (n+(n-1))^2