Problem Statement- King Arthur sets up a game with all of his knights. Lets say there are ten knights, the all sit in a seat. King Arthur will then start with knight one, saying he is still in the game, then going to knight two, saying he is out of the game. He does this until there is one knight who is left, the winner of the game. Our task was to find a formula that will be able to determine which knight is in the winning seat before the game starts.
Process- The first thing that my group did when we heard the problem was list out numbers and do the same in and out sequence that king Arthur did. We found that we all did really similar things, we drew a circle with numbers listed on the outside, crossing off numbers until there's one last seat. We started with numbers 1-10, then 1-15 and so on; trying to see of there was a pattern. We found out that the first seats that would be out of the game were those sitting in an even number seat. My group was lost for a really long time, we did not know how to create an equation that will pre-determine which is the winning seat, so we kept on using the method, going up to seat numbers like, 36, 45, 50 and even 100 seats!
Solution- While our group was stuck, there was another group that was able to create a formula that worked in finding the winning seat for any number. the equation is y=k-2^p·2+1 . K= the number of nights
p= two to the power of__. with 2 to the power of p, we plugged in numbers that give us powers of two closest to K.
for example: if K= 36, we would use 2 to the power of 5,which gives us 32. (we can't use 2 to the power of 4 because its too low from 36, and 2 to the power of 6 is 64, which is too high.
if K= 36 we use the 2 to the power of 5, (which gives us 32) and subtract 32 from 36 leaving us with, 4, then we multiply 4 by two, giving us ,8 then adding 1 (because we know that the winning seat can't be an even number) making the winning seat number 9.
This formula works because it works for every number, odd or even. MY group would use the formula and have one person use the in and out sequence with K and we checked to see if it worked, and it did.
Evaluation/ reflection- Something that pushed both myself and my groups thinking was trying to create a formula that works for wvery number. I think that this time around with solving a problem with a group, I feel that everyone was able to be on the same pace and take the time to try and understand each other during the group quiz. With this group, there was someone who is usually quiet and doesnt get involved but that person was able to help when we divided up the work and understand how we got the equation, (even if it was wrong in the end)
I think i can recognie when I need help on a math problem but I dont have the confiencence to ask why something is supposed to work a certain way, I'll ask how to use it , but I wont be able to explain it to others who ask me for help.
With this problem, during thr group quiz I feel like I took on a sort of "leader" role because I was the first person to have an equation and pose it to the group, which worked for every number except 2 :( I used the original equation y=k-2^p·2+1 and took away the plus one part (since the problem was asking what would an equation look like if seat number one was out first) so it was y=k-2^p·2 which I was able to explain to my group and they were able to explain it to each other.
I think for this problem I would give myself an A because I think that my group made it really easy to share ideas and we were all able to ask and answer questions that we all had. I think I deserve an A because when the quiet person in the group looked lost I re directed them to the problem and asked them if they understood and if they wanted me to explain how I got that equation, and then they were able to explain it in there own words.