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Suppose you have 3kg flour, and you are asked to divide it to three 1kg parts using a balance scale.

It seems to me that it's impossible to do with a finite number of weighing, but I can't see how to prove it. Is this a known problem? Any hint is appreciated.

Please comment if the question is vague or needs further clarification.

Edit: I'd appreciate it if you formulate this problem with reasonable assumptions into a pure math problem.

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  • $\begingroup$ That is the well known problem of Trisection of an angle. $\endgroup$ – hamam_Abdallah Sep 28 '18 at 19:27
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    $\begingroup$ @Salahamam_Fatima How is it related to trisection of an angle? Trisection of an angle is impossible with an unmarked straightedge and compass. What does this have to do with weighing flour? $\endgroup$ – saulspatz Sep 28 '18 at 19:29
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    $\begingroup$ I wonder how we are even able to divide the flour into two equal parts in a finite number of steps, given a mathematically perfect balance scale and the ability to transfer arbitrarily small (but unmeasured) quantities of flour from one pan to the other. Perhaps you simply assume that division in two equal parts can be done in one step, but then you should state that assumption explicitly. What are the fundamental flour-weighing operations from which we can build strategies for dividing flour in a finite number of weighings? $\endgroup$ – David K Sep 28 '18 at 20:54
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    $\begingroup$ @David K Good comment. I think the author's intent was to disguise the trisection of an angle problem into a flour weighing problem. But if the assumption is an ability to divide the flour into two equal parts with a finite number of trial and error weighings then my solution below requires no more than that. Equal parts from A and B determined by balancing the scale can be added to a smaller C quantity until A balances with C. $\endgroup$ – Phil H Sep 30 '18 at 0:31
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    $\begingroup$ Put a known 1 kg weight on one side of the balance, and weigh out enough flour to balance it. Put that flour aside and weigh out enough flour from the remaining part to balance that. Weigh the remaining flour and see how close it comes to 1 kg. If you can't tell a problem then you did it right. If it doesn't match, maybe you weighed something wrong or maybe the original flour was not exactly 3 kg. So which do you trust, your standard 1 kg weight or the 3 kg flour sample? $\endgroup$ – J Thomas Oct 1 '18 at 10:56
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All you can do with your scales is: divide by $2$; add two numbers; and subtract two numbers. And you can't get to $\frac13$ from $1$ with a finite sequence of these operations, because all numbers created by them have a denominator that is a power of $2$.

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  • $\begingroup$ Your answer makes sense, but I have difficulties to completely relate it to the problem. Can you formulate the original problem in terms of a pure math problem? $\endgroup$ – Asmani Sep 29 '18 at 17:01
  • $\begingroup$ What if we completely avoid dividing by 2, or multiply by 2 when needed? $\endgroup$ – Asmani Sep 29 '18 at 17:09
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There are two possibilities for this answer which depend on two different sets of assumptions. The first is that this is strictly a theoretical math problem, similar but not identical to the trisection of an angle problem, whereby dividing a quantity of flour into two halves any number of times cannot result in three equal quantities. This also assumes that the flour is an infinitely divisible substance and not a finite number of discreet grains. A series of divisions $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}......, \frac{1}{2^n}$ can never partially sum to $\frac{1}{3}$ for any finite number of divisions, as shown in the previous answer by TonyK.

The second assumes a more practical approach which considers the trial and error aspect of balancing two quantities of flour. It also allows for an imprecise division into two parts whereby individual grains of flour are not themselves divisible. Divide the flour roughly into $3$ equal piles A,B and C. Put A and B on the scale and adjust until they balance. Put A and C on the scale and adjust a lighter C with equal parts of A and B or equal division of excess C to A and B (both determined by removing quantities A and C and determining the equal adjustments with the balance) . Repeat until A and C balance. Then all $3$ will be equal within the accuracy of the balance beam. I certainly don't see this as impossible.

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  • $\begingroup$ Yes, I get down voted for a practical rather than a theoretical mathematical perspective. $\endgroup$ – Phil H Sep 28 '18 at 22:38
  • $\begingroup$ Sir, I feel instead of seeing it as a math problem, thinking out of the box in the way as you did, suits the purpose. $\endgroup$ – Narendra Deconda Oct 1 '18 at 11:05
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    $\begingroup$ +1 for clarifying assumptions. One cavil. The mathematical problem is not equivalent to the angle trisection problem. It's equivalent to trying to trisect just using bisection, but it takes more work to show trisection can't be done with Euclidean tools. $\endgroup$ – Ethan Bolker Oct 1 '18 at 16:16
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    $\begingroup$ @Ethan Bolker Yes, you are correct, it's a little more complicated with Euclidean tools as one can construct an angle of 30 deg which is 1/3 of a right angle. Looking it up, it reduces to an algebraic problem of finding roots of a cubic polynomial. $\endgroup$ – Phil H Oct 1 '18 at 16:34
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Theoretically, assuming it is possible to use the balance scale to divide any flour quantity to two equal parts. Make a continuous sets of two half dividing rounds. At end of each set, sum one side and use the other side to continue the two dividing rounds sets. This shloud give you the sum: $$ \color{red}{S} =\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\cdots =\sum_{n=1}^{\infty}\frac{3}{4^n} =\color{red}{1} $$

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  • $\begingroup$ Clever, but this is an infinite sum with infinite weighings where the requirement was for finite weighings. $\endgroup$ – Phil H Oct 2 '18 at 21:47
  • $\begingroup$ @PhilH Thanks Phil. As you said in your creative answer “thinking out of the box” ($+1$ from me $2$ U). Meanwhile, I noticed the OP is more interesting in a solid argument confirming his opinion regarding the impossibility of solving the puzzle. The question did not mention anything about the number of weightings (like with minimum number of weightings or with less-than say $100$ weightings). Of-course, in practical it should be a finite number. Nevertheless, theoretically it is solvable as explained. Thanks again & good-luck. $\endgroup$ – Hazem Orabi Oct 3 '18 at 0:21
  • $\begingroup$ An ambiguous but interesting question nevertheless. Your answer is useful. Thanks Hazem $\endgroup$ – Phil H Oct 3 '18 at 0:37
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This seems rather easy to solve. Call the 3 quantities A, B, and C. First, get A and B to balance and compare either one of those to C. If C is heavier then attempt to estimate the overage D and put half of D into A and the rest into B. If C is lighter than A (or B) then take a similar amount out of A and B and put in C. Just repeat this process until the balance cannot detect any difference between A, B, and C.

I know this is not very mathematical but it makes sense to me so it is a valid answer.

Also I doubt you will ever get this exact since some flour will likely stick to the container you are using to hold the flour.

But hey a half ass solution is better than no solution!

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