(Sept 28) Balancing the Scales
To solve this problem we must start by going through the cases. First 1 gram of herbs. To measure this we obviously are going to need a 1 gram weight. Then 2 grams. To measure this We can either use 2 one gram weights or if we put the 1 gram weight with the 2 grams of herb we can get use a new weight that is 3 grams. So we have 2 distinct weights. Now 3 grams we can use the weight we have. For four we can use the 3 and 1. For 5 we do not want another 1 gram weight, so if we put both weights with the 5 grams of herb, we can get a new 9 gram distinct weight. Now skipping some steps the next number we get stuck on is 14. But we only have one weight left so we need to be able to reach 40 grams with the remaining weight, so it must be 27g. Can we get 14 with this weight? If we put all 13g of weight with the 14g of herbs we can balance it with our 27g of weight. And the rest of the weights fall into place as weights can be placed on both sides. We can see that the pattern is 3^0, 3^1, 3^2, 3^3, but why? It must have something to do with how each weight has 3 paths. Leave it off, or place it on either scale which adds or subtracts weight. So a 5 weight one pan system therefore must have a similar pattern. One pan scale reduces the choice of each weight by one so each weight can only be put on or left off of the scale. So this should be in powers of base 2. 2^0, 2^1, 2^2, 2^3, 2^4. Or 1, 2, 4, 8, 16. This system can only weigh up to 31 grams as that's what it adds up to, so if we want to weigh more we would need to add a 32g weight. Checking a random number 25g of herbs we can see that this is just 16+8+1. Checking all steps this does work out. This ties directly into how ancient Egyptians used doubling to solve multiplication problems. Where they would just double numbers and add the results which is what binary decomposition is in the present day. We can extend this problem by creating a new problem with 4 choices and now students can see the pattern clear as day! Where the number of choices changes the base of your solution. Doing this type of "long" mathematics will actually help students take detours to find solutions. In schools, we are constantly taught one way or the most efficient way to solve a problem. For example, memorizing multiplication tables, or calculating derivatives with the "rule". We forget what the true meaning of multiplication is and just take it at face value. Before we move on to shortcuts we MUST understand the foundations so that when the "trick" doesn't work we can find a multitude of different ways to find the solution. That's what mathematics is about, finding different ways to the same solution.
Comments
Post a Comment