The problem of dividing 3kg flour to three 1kg parts using a balance 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.
 A: 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.
A: 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:
$$ \color{red}{S} =\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\cdots =\sum_{n=1}^{\infty}\frac{3}{4^n} =\color{red}{1} $$
A: 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 $2$.
A: 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!
