This problem is about a king who is trying to find a way to verify different weights, using the least amount of weights as possible. There were two different ways we could go about this. One of those was only putting weights on one side to equal whatever item you’re weighing. The second method was putting weights on both sides to find a balance. The problem was asking us to find a solution for each method that used the least amount of weights as possible but still covered all the numbers below them.

For the first method of only adding weights to one side of the scale, we split the problem up depending on how many weights we had. When we only had one weight, we found that the 1 weight, obviously, is the only one that covered all of the numbers below it. When we had two weights, we needed a 1lb and a 2lb weight. This combination gave us 1lb, 2lbs, and 3lbs. When there were 3 weights, we found that having a 1lb, 2lb, and 3lb weight gave us the solution we needed. With this combination, we covered the weights 1 through 6. At this point, we started noticing a pattern that the lowest numbers gave you the most options. When there were 4 weights, the best option was weights 1-4, because they covered weights 1-10.

For the second method of adding weights to both sides, we also split up the problem based on the amount of weights that we were allowed to use. When we were only allowed one weight, again we went with the 1lb weight. However, with this method we could eliminate the need for a 2lb and 3 lb weight. To be able to count as many as possible, we decided that having a 1lb weight, a 4lb weight, and a 6 lb weight was enough to get us all of the numbers up to 7, apart from 10 and 11. We chose 4 because it was the next lowest number and we chose 6 because it was 2 away from 4. With this 2lb difference, we could make up for not having a 2lb weight. And if you added the 1lb and the 6lb, that was a 3lb difference from the 4lb weight. With this combination we made up for not having a 2 and 3 lb weight and we were able to get higher numbers.

For the second method of adding weights to both sides, we also split up the problem based on the amount of weights that we were allowed to use. When we were only allowed one weight, again we went with the 1lb weight. However, with this method we could eliminate the need for a 2lb and 3 lb weight. To be able to count as many as possible, we decided that having a 1lb weight, a 4lb weight, and a 6 lb weight was enough to get us all of the numbers up to 7, apart from 10 and 11. We chose 4 because it was the next lowest number and we chose 6 because it was 2 away from 4. With this 2lb difference, we could make up for not having a 2lb weight. And if you added the 1lb and the 6lb, that was a 3lb difference from the 4lb weight. With this combination we made up for not having a 2 and 3 lb weight and we were able to get higher numbers.

For method one, we discovered that buying base 2 weights and one base 1 weight would be the best weights for the king to buy and be able to balance any packages. For method two I did the same thing and discovered that base three weights are the best for the king to buy along with a base 1 weight.

For me, this was definitely one of the most challenging, but entertaining, problems. The math itself was not complicated at all, the pattern/evaluation thinking style required to solve this problem was what was challenging. For something this repetitive, an algorithm would have been the best way to solve this.

I think I did a good job on this problem because we found a solution relatively quickly. I also find it necessary to give Nayeli a shoutout because we work really well together, and we will be going to Dartmouth together.

]]>This problem asked how many different ways there are to arrange tiles of 1 inch by 2 inches in an area of 20 inches by 2 inch.

For this problem, we started smaller to see if we could find a pattern and go from there. We started with a 2 inch wide by 4 inch long path since the problem gave the example of a 2 inch by 3 inch path. We set up a x and y table and kept making the path longer and seeing how many ways there were to rearrange it. As we looked back at the table, we realized that the numbers started looking like a very familiar pattern, the fibonacci sequence. We then realized that the length of the path would be the number of term in the sequence. We then predicted the next number and tested it out.

For our solution we got 10,946. We got this by finding the 20th term in the sequence.

At first I thought this problem was going to be very easy and that I could just find the number of times I could rearrange them in a 4 inch long path then multiply it by 5 since you could do that 20 times. Then once I started doing the x and y table it wasn’t that simple. Drawing out the different arrangements for the different lengths was also hard because I was scared that I missed one and I actually repeated 2 one time which kind of threw me off at first.

I think I did good in this problem even though it took me a little longer to find the answer. I kept trying different ways that I thought would work and wouldn’t in the end but I didn’t give up. I persevered through until I found the answer. I also feel like I was pretty creative in the way I found the answer, I kept looking for different ways since my initial ideas didn’t work. In the end, I came across the answer kind of by accident but I used my prior knowledge to make the x and y table and saw the pattern in that way.

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This problem asked what number of moves it took to kill 5 switches. But there were very specific rules. You could just switch a switch if the change directly to the left was on and each other switch to one side was off. The change most remote to one side didn't make a difference and could be switcher paying little respect to what position the greater part of alternate switches were in. At that point it asks what number of moves it would take to kill 10 changes lastly to demonstrate any equation or example I have thought of.

To start with I began small, I began with 2 switches and what number of moves it would take to turn every one of them off, then I proceeded onward to 3 and 4 until I achieved 5. When I had done each of the 4, I made a x and y table to attempt to see an example. I saw that 3 switches required 3 moves more than 2 switches, then that 4 switches required 5 moves more than 3 switches lastly 5 switches required 11 moves more than 4 switches. At first I thought it would increment by odd numbers which would make 6 switches require 13 moves more than 5 switches, however that hypothesis didn't work out subsequent to the base number of moves for 6 switches was 42. At that point I saw that at times it was multiplying the quantity of moves before and at times it was one more than twofold of what was some time recently. I utilized this example to make sense of what number of moves it would take to kill 10 switches.

The minimum number of moves required to turn off 5 switches is 21. The minimum number of moves required for 2 switches is 2, for 3 it’s 5, for 4 it’s 10. I couldn’t find a general formula for all cases, only found separate formulas for odd number or even number of switches. For odd numbered switches, you double the number of moves from the number of switches before and add one. For even numbered switches, you simply double the number of moves from the number of switches before.

I enjoyed this issue despite the fact that I couldn't construct a general formula. This helped me to remember the tower of hanoi and a problem we did in Mr. B's class junior year. I don't recollect precisely how it went, but there was a line of 4 individuals taking a seat and you needed to get the last individual to be the one and only standing up yet a man could just move if the individual directly before them was standing up and others before them was taking a seat. Fundamentally the same as the switches. This problem was a good reference because between each move, the principal individual, or in this present issue's case, switch, would move.

I think I did sort of good on this problem. I was not able to figure out the general equation yet I think I got quite close. I got innovative without an equation to locate the base number of moves it would take to kill 10 switches without really working it out on the grounds that it would have been a ton of moves. I invested a considerable measure of energy in this problem, attempting to locate the base number of moves since I would dependably get stirred up and botch up. The first occasion when I couldn't discover any examples in my number so I backtracked and attempted again and got a littler number so I checked the majority of my base number of moves.

]]>This problem is around a mother with 2 kids who stroll by a gumball machine and both her children need a gumball of the same shading. The problem had three sections to it. The first being what is the most extreme sum the mother would need to spend all together for her children to have the same shading gumballs. The second part inquired as to whether there were more hues or more children? What's more, the last part inquired as to whether now the children needed diverse hues?

For this problem I began with the 2 children and 2 colors then proceeded onward to 2 children and 3 colors and continued changing the numbers while making table. In any case, this table was not quite the same as the past ones since there is a third variable in this problem. When I had different circumstances, I found the example in my table and found an equation. Once I understood why the equation worked, I went onto the following part where there are just 20 gumballs in the machine and it lets you know what number of number of every shading there are. This was somewhat different on the grounds that the past circumstance was accepting there was a 50/50 shot of getting either shading however with this new one there was not a 50/50 chance furthermore now the children need distinctive hued gumballs.

Our solution for the first part was the equation c(k-1)+1=p. “c” being the amount of colors of gumballs, “k” being the amount of kids, and “p” being the pennies used. For the next part where the kids want one gumball each of different color, we got 9 since there are 8 yellow gumballs, 7 red, and 5 black, if by chance, you kept getting the same color gumballs, by the time you got to the 9th gumball, there is no way it could be the same color as the 8 before since there are no more than 8 gumballs of the same color. We used the same process for when the twins each wanted 2 gumballs of the same color but different from each other. If you were to get the first 8 gumballs the same color, the 9th gumball would be a different color, then if the next gumball happens to be the third color you have not gotten yet, the next gumball (11th) has to be the same color as any gumball you’ve already gotten. Therefore the maximum the mom would spend would be 11 pennies.

This problem was hard to grasp because we did not know what to do if we did not know how many gumballs were in the gumball machine in total. If there were 400 gumballs in the machine, and half are one color and half the other, even though it’s nearly impossible, it could happen that you get 200 gumballs of the same color, then it would take you 201 pennies in order to get different colored marbles. Then we realized that it didn’t matter since we were trying to get them the same color marbles.

I am proud of myself for completing this problem, because it is definitely one of the ones in which I struggled the most out of all of these. I relied heavily on my partners when finding a solution, but am proud of us for figuring it out on our own without asking for help from teachers.

]]>In this problem we were trying to find how long it would take to move 64 disks from one peg to another, assuming you moved at a rate of one disk per second. The disks are originally stacked on the peg to the left, in ascending order top to bottom. The rules are you can only move one disk at a time and you can’t place a larger disk on top of a smaller disk.

When we first started this problem, Nayeli told us that she had posed this question to her brother brother as they were driving home from school one day. Surprisingly, they were able to come up with an answer using only our phones(they would ask siri to multiply for us because the numbers were so big). But we had to restart all of her work because she did not remember where she had started. When we wrote everything down, we were able to summarize the expanding difference between the number of moves it took to win based on the number of disks. This was, when there were 2 tiles, the next time it would take 2^2 more moves next time to win. And then when there were 3 tiles, the next time it would take 2^3 more moves next time to win. This could be summarized by the equation 2^D=M, where D represents the number of disks and M is the number of moves. To solve this problem, we simply substituted 64 where we had the number of disks.

When you take 2 to the 64th power, you get a huge number, which is the amount of moves you’ll have to make. To convert this into a more reasonable set of time, we turned it into years and got 584,942,417,355 years.

This problem was relatively simple. The pattern that I found was pretty straightforward. I think this problem is more of a warm up than a problem of the week. The solution is straightforward and simple, so this is definitely a problem I would do with the rest of the class just for fun.

I think I did very well on this problem, and believe it was an easy solution for me to find because I was able to relate it to other problems that I had previously solved in classes like challenge math and CS50.

]]>The “One Mile at a Time” problem asked us to solve for the speed of an individual under certain circumstances. First, it asked us for the speed when, after an hour of their trip, the travelers saw a milepost with a two digit number. One hour later, they saw another milepost with the same two digits but backwards. After another hour, they saw another milepost with a three digit number using the same milepost as the first milepost except with a zero between the two digits.

We started by trying random numbers like 28 but then realized we should go in an order to see if we could find a pattern. we started with 01 and went in increasing order. We would write out what the numbers would be for each milepost and then figure out the miles traveled between each milepost. We knew the mileage had to be the same between the first and second milepost as the mileage between the second and third milepost. We started realizing that the mileage between each milepost was always a multiple of 9 but didn’t understand how that tied in. When finding the algebraic expression, we remembered that each distance traveled was a multiple of 9 and tried to include the number 9 in my algebraic terms. We saw that 16 was pretty close to 18, a multiple of nine so we set up 16 as (9*2)-2 and tried something similar for the next to mileposts. We realized they were both 2 miles short of a multiple of nine and went on to expand 61 to (9*5)-2 and 106 to (9*12)-2. Then we saw the pattern that each time you are multiplying 9 by 5 more than the last. |

As we were listing the numbers, we reached 16 making the second milepost 61 and the third milepost 106. They each had a distance of 45 between each milepost. Our answer for the algebraic terms is

9(2+(5(P-1)))-2=M. “P” being the post number, and “M” being the miles on the post.

9(2+(5(P-1)))-2=M. “P” being the post number, and “M” being the miles on the post.

For this problem we mainly used hypothesizing and experimenting, because we hypothesized that if we kept with my pattern, eventually some numbers would match up and then experimented with the numbers. We also did a lot of experimenting when we were trying to find an algebraic explanation for the solution.

I thought this was a really hard problem to start because when I tried the first number, I thought that it wasn’t possible. But once I started doing it in order, I realized that it was just a matter of time until one of the numbers matched up. I was really confused with the second part of explaining it in algebraic terms because there are dozens of ways to represent any number using different equations and didn’t think I could find one that could help me see something in common between the numbers on each milepost.

]]> We were presented with the following situation: there is a game of spinners where all three players spin their spinners and the one who lands on the highest number wins. The first part was to compare the spinners and see who's spinner was most likely to win. The second part of this problem was to try and create a new set of spinners to see if you could recreate the same pattern as the original set of spinners. |

For this problem, I used skills that Ms. Chen taught me in sophomore year. I made a probability tree to compare the chances of one person winning over the other. After I got my results, I looked for patterns in the wheels to try and figure out why the answer was so interesting.

## My SolutionFor part one, I found that there was no one person who had a better spinner. I learned this is called a non-transitive relationship. To re-create a set of spinners that could also have non-transitive relationships, I took a close look at the different sections of the spinners. I split them up and compared them based on their location, as you can see in the picture to the right. | |

Although I was unable to recreate a different set of spinners, I did find a pattern. Each section on the different spinners had a set of consecutive numbers. Each spinner had one section with the highest number, one section with the middle number and one section with the lowest number.

I thought the answer was really interesting because you would assume that in situations like these, a<b<c, but the answer was something more like a<b, b<c, and c<a. This relationship is called non-transitive, compared to the more straightforward transitive relationship (a<b<c). I thought this was a really interesting question and I think this would be a good challenge math problem. I will definitely remember non-transitivity because it came as a surprise to me. I also am not really sure how to recreate the spinner. I noticed a pattern but as hard as I tried, I couldn't make a different set of spinners.

With this problem, I grew a lot in my hypothesizing and experimenting abilities. I spent a lot of time trying to create a new and unique set of spinners that also had a non-transitive relationship. Although I was unsuccessful, I still think that persevering helped me develop my experimenting skills. I was looking for patterns in the spinners, but I couldn't replicate the pattern that I thought I was seeing. If there had been another pair of spinners, I think I could've maybe created another spinner. It's difficult to see patterns when you only have one example.

Even though I didn't find a different set of spinners, (starting to think there aren't any) I think that I still learned from this exercise. I got to explore transitive relationships and probability. I will definitely think about this in the future when it comes to probability and determining my odds or what the probability is of things coming out in my favor. Next time that I'm calculating my chances, I will remember that there is not always going to be one superior option.

]]>Even though I didn't find a different set of spinners, (starting to think there aren't any) I think that I still learned from this exercise. I got to explore transitive relationships and probability. I will definitely think about this in the future when it comes to probability and determining my odds or what the probability is of things coming out in my favor. Next time that I'm calculating my chances, I will remember that there is not always going to be one superior option.