# How many weights. This discussion has hints in the last section. If you don't want to see them, stop when you see the title **The Hint** You have a balance scale. You need to be able to determine the weight of any sample between 1 and 100 grams. The samples are all in positive integer quanta, i.e. 1, 2. 10, 27, 30, 55, ... but no 3.5, 5.7 or $\pi$). Likewise, all weights are positive integer multiples of 1 gram. Call the pan on which you put the sample the _sample pan_, and the other the _balance pan_ You are allowed to put weights on either the _sample pan_, or the _balance pan_. What are the minimum number of weights required, and how many of each are required? By this I mean **both** what I call the 'denomination' (say ... 1, 7, 9, 12, 35, ... ) which are the actual weights **and** how many of each you need to add something up. This is sort of like a vector space, with the individual weights as the 'basis vector' and the number one multiplies it by as the 'coefficient' in a sum. ## Analysis. Let's start with a simpler problem: ## You can only put weights on the balance side of the balance. Again, when you hit the title **The Hint**, skip it if you don't want to read the hint. I don't give the solution, you still have to do the work, but I think you'll be pretty clear what to do when after you've read that section. In this case, you're just adding weights up in the normal way. By this I mean something like... You put a sample of 32 grams on the _sample side_, and you put weights on the _balance side_ until it balances. That is a **summation process**. Here's one way that could happen. $32 = 3*10 + 2*1 = 3*10^1 + 2*10^0$ where we're working in base 10 arithmetic.[^5] Now, you could say : wait, you could also do: $32 = 6*5 + 2*1$ and that's fine, but it muddies the math a bit, because the number '32' **means** $3*10+2*1$. If you want to work in a different base (base 5 here), you should use the representation of the number thats appropriate to that base. In base 5 it would be $32_{10} = 6*5^1 + 2*5^0 = 62_{5}$ The subscript says what base we're working in. Or, we could do this sum in base 7. $32_{10} = 4*7 + 4*1 = 4*7^1 + 4*7^0$ [^4] $44_{7} = 4*7 + 4*1 = 4*7^1 + 4*7^0 = 32_{10}$ The point is: the way we write numbers down is just a shorthand representation of a process of summation using powers of a base multiplied by appropriate coefficients in the range of [0 - base-1]. The number '1023' is just shorthand for $1*10^3 + 0*10^2 + 2*10^1 + 3*10^0$ If no one indicates otherwise, base 10 is assumed. So, with those two examples I'm just showing that I can write the number $32_{10}$ in different bases. Sure, they have different digits, but it's still the same quantity. They're **different representations** of the same quantity. ## What about putting weights on the _sample side_ The problem says you can put weights on either side of the balance scale. In this section I'm going to discuss the effect of doing that. The easiest way is to use an example. So, to represent $99_{10}$, we could do... $99_{10} = 9*10^1 + 9*10^0$ But, we can also do $99_{10} = 1*10^2 - 1*10^0$ In the first case, we'd need 9 weights of 10 grams and 9 weights of 1 gram. In the second we need 1 weight of 100 grams and one weight of 1 gram. So it's more efficient in terms of the number of weights (1,1) versus (9,9), but the same in terms of the number of denominations: both need 2, though different _denominations_: (1,10) versus (1,100). But, we can't subtract anything by putting it on the balance side_. Putting something on the *balance side* only results in an **addition**. But we can move that weight to the _sample side_, that is, by **adding** that weight to the sample! $99_{10} + 1*10^0 = 100_{10} = 9*10^2$ So, putting something on the sample side is equivalent to **subtracting it from the balance side**. ## The hint. **spoiler alert** Stop here if you think you have enough to solve the problem! I'll just put a picture here so that you won't see the next line without scrolling. ![[Jac-and-Rou.png]] While working in a particular base, the most efficient way to represent a number (both in the number of denominations and the number of those denominations required) is by summing powers of some base. $b^0, b^1, b^2,\ldots$ and so on, where the coefficients are the set [0,1,2,...,b-1]. Clearly, in our problem that requires the physical presence of objects, in order to be able to sum to any number in a range, you'll need **both** the necessary set of powers of a base _b_, **and** you will need _b-1_ of each base to serve as the coefficient for each term (that is, 9*10 = 10 + 10 + ... + 10, (nine times). Same-same for the other powers of the base) Now, I don't know this for a mathematical fact, but I think you can construct a 'basis' for all the numbers in your range of interest using all sorts of sets of numbers, provided they 'cover' the range[^1], but the smallest set of numbers that will 'cover' that range will be the powers of some base.[^6] We universally use base 10 in our everyday lives, but computers (and nerds) use base 2 and base 16, and the Babylonians used base 60![^2] Which base you use is only important (if it's important at all) when solving a particular problem, where a particular base is suggested by the 'shape' of the problem. Otherwise, the choice is merely a convention. In the problem about weights, the *number* of each weight we'd need is whatever the base is minus 1 (a coefficient of 0 just means you don't use that power). If you use base 10, you'll need 9 of each weight to be sure you can generate all sums in the range of interest. If you're in base 7, you'll need 6 of each weight. And you'll need whatever powers of that base minimally cover the range. For example, in the case of 99 above, clearly we'd need the set powers of 10 to be [1,10,100], and for the coefficients we'd need [1-9,1-9,1], that is, 9 weights = 1, 9 weights = 10 and 1 = 100, for a total of 19 weights.[^7] Oh, one last thing. I misspoke just a bit when I said the coefficients are the set [0,1,2,...,b]. That's the way it's always taught, because that's the way we do it in our normal day-to-day life. But the set of coefficients could be [-b/2,-b/2+1,...,0,1,2,...,b/2] where the division is **integer division**, where you discard any remainder: e.g. 7/2=3. So, in base 7 the coefficients could be [0,1,2,3,4,5,6], or [-3,-2,-1,0,1,2,3]. In base 5, you'd need [0,1,2,3,4] or [-2,-1,0,1,2]. Frankly, it could be [-1,0,1,2,3] in base 5. So long as the set of coefficients has [base] elements, it contains 0, and no repetitions, it will work.[^3]. Who knows, may be there are situations that suggest using base 5 with the coefficients [-1,0,1,2,3]. Strangely enough, in base 10, the coefficients are nominally [0,1,2,...,9], but, because 10 is divisible by 2, the equivalent set of coefficients that go into the negative integers has to include 5 on one side, e.g. [-4,-3,-2,-1,0,1,2,3,4,5] or [-5,-4,-3,-2,-1,0,1,2,3,4]. Just an oddity I'd never thought about that I thought I'd mention. Given this problem, do these considerations suggest a solution? # Busy Bee Challenge https://youtu.be/AKiA7JBgJdQ?si=_Tdj3bxDSf3_1_Tp # Anning-Erdos Geometry Problem Are there infinitely points in the plane that are integral distances from each other https://www.youtube.com/watch?v=fpIc-FE4c5U The started with a simple problem: Are there infinitely points in a plane so that the distances between each pair of points is a rational number not integral)? - [ ] Start by creating line and a point not on the line at right angles, then extend the line to the right, making a series of right triangles. Can we find points on that line such that 1) the points on the line are rational numbers and 2) the hypotenuse formed by connecting the point to the 1 point not on the line is a rational number. # Footnotes [^1]: The set of coefficients might be a problem, I haven't really thought about it, and I'm unlikely to anyway, since it's a basic theorem that in any base, the powers of the base are the most efficient 'basis' [^2]: Think about memorizing the Babylonian multiplication table! **yikes**! [^3]: I think! [^4]: (so, 4 '7s' and 4 '1's) [^5]: So, in this case you'd need 3 weights weighing 10 (that is 3 weights of the '10' denomination) and 2 of the 1 denomination. [^6]: By this I mean that you can generate some set of numbers in some base, say base 10 to keep it simple, that, along with the coefficients for base 10, will 'cover' the range, but you'll find out that you'll need more 'denominations', more 'basis vectors' than if you just used the powers of 10. [^7]: That is, if we're using the convention of the coefficients of [0,1,...,9]