When we were first introduced to this problem, we were told to reflect for 5 minutes questions or ideas we could do to start this problem. As a group we shared our ideas and agreed on drawing it out to visualize it, and finding a pattern based on making a table; k for knights, and w for the winning chair.
We realized that the winning chair was 1 by each power of 2, so we knew the equation we were suppose to come up with had to have something to do with this rule. After a while of being stuck, we asked our teacher to help us out a bit because we were having problems with writing an equation for it. Time ran out and we didn't get to solve this on our own, and a group who came up with one shared and explained to the class. The next day we tested out this problem in real life. Both classes got there chairs and took them down to the field and we set up in a circle. Since we knew the equation many students were trying to sit in the correct spot to win by doing the math with the equation. |
The solution is subtracting the number knights by the nearest creates integer of base 2 exponent. After that multiply by 2 and add 1 for the winning seat.
Another group shared their equation with the class that can find the winning seat for as many knights. Y=(K-2^P)2+1 K, represents the number of knights/chairs in the table. P, being the power of 2; closest to the power of k. This works because it would give us the right answers for odds or evens. We didn't have to do the steps of having to write them out and crossing them one by one. |