Problem Statement
In this problem the main rule that you are given is the "In" and "Out". So say there is a total of 12 knights and to narrow it down to 1 knight, King Arthur goes around the table saying knight #1 is in and Knight #2 is out and so on. When King Arthur is finished with the first round that ended #12 being out, he will continue with the same pattern without resetting. So the problem is, if you know the total number of knights, doesn't matter the total, which number would be the last one standing?
Process
The first few attempts at solving this, we created at-chart with x being total #'s of knights and y being the knight that won. One thing that we noticed straight away is that all even numbers are removed on the first round so its impossible that the last knight will be even. We started with different numbers and worked from that. We then realized that it resets every powers of 2 like 2^2, which is total of 4 knights and the knight that will be the last is 1. For the numbers that don't reset to 1 go from the odd order like 3.5.7.9.
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Solution
The formula that we came up with was S=(k-b^p)2+1. For K-b^p, K represents the total number of knights and b^p represents the nearest power to the total number of knights. With subtracting both of those we get to see the difference from the reset number being one to the total number and which seat is the winner. The reason why it is added by one is because the number has to be a odd number because all of the even numbers were taken out.
Ex.
k=68
The nearest power to 68 is 2^6 being 64 and there is no other power inside of 68. So far you have:
S=(68-2^6)2+1
S=(68-64)2+1
S=(4)2+1
S=9
Ex.
k=68
The nearest power to 68 is 2^6 being 64 and there is no other power inside of 68. So far you have:
S=(68-2^6)2+1
S=(68-64)2+1
S=(4)2+1
S=9
Reflection/Evaluation
Something that pushed my thinking was to make it into a formula because i struggle with that. But with help from my group mates I was able to understand how the pieces fit together and how to make a formula that works with this type of problem. Something else that we struggled with was to involve others in our group to share ideas. We did work with each other and helped each other but we did our work individually. I am not afraid to ask for help when I am struggling and I do recognize when I need help. But before I do that I do try to see if there is any way that I can solve it on my own before I ask someone else. When I help someone else, I try to explain it to when they can understand and help me in the process too. When it came to the group quiz, I think that is when we all contributed the most in because we faced a few challenges where the formulas weren't correct and we had to work together to see if there is a way to solve it. We all talked about the patterns that we saw and how that can help us. If I were to grade myself on this problem, I think I would give myself a A because