This puzzle connects with the idea of different bases, as the two-pan scale connects to the base-3 system, given that we have 3 choices of where the weight can be placed (on the pan with the herb, without the herb, or not on the scale at all). Since base 3 then uses values which connect to the powers of 3, we see that we have weights that are equal to 1, 3, 9, and 27 aka 3^0, 3^1, 3^2, and 3^3, where within base 3, all numbers from 1 to 40 can thus be written as some linear combination of these values. Since the one-pan scale only gives us two options of placement for the weight (on or off of the scale), we instead use the base-2 system to hit every value from 1 to 31 using linear combinations of the values 1, 2, 4, 8, 16, which corresponds to 2 to the powers 0 through 4.
To extend this puzzle, one could modify it to allow for 2 weights of the same size on the one-pan scale and see how this then works/forces a base-3 system. For example, seeing that to get a value like 20, you'd use two 9g weights and two 1g weights, whereas once you upgrade to two-pan scale you can get the original setup where you have 4 completely different weights only.
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