## Problem Statement

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.

## Process

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.

## Solution

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.

## Problem Evaluation

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.

## Self Evaluation

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.