Generalization of the two bucket puzzle

Simon's answer points out that the Euclidean algorithm shows that gcd(A,B) divides C is necessary, but the lack of large container makes the problem more difficult, because obviously you can't get $C$ if $C \gt A+B$. However, the following modification of the algorithm seems to work.

Let's assume $A \lt B$ and gcd$(A, B)=1$ for simplicity. By pouring from B into A and dumping A, you can get any positive integer $B-nA$ left in B. Do this until the answer is less than $A$. Then by transferring the contents to bucket A and filling it from B into it, we get $2B-(n+1)A$. Then subtract $A$ again until you have $0 \lt 2B-kA \lt A$, and we can iterate this process to get $3B-(k+1)A$, etc. This gives any integer linear combination $rB-sA$ up to the size of $A+B$ because once you get the right multiple of $B$ into the combination, you can always add bucketsful of $A$.

You just need to get $rB\equiv C$ (mod A) in order to find a combination for $C$, which happens if gcd$(A, B)=1$. With this algorithm run in the general case you can get any multiple of gcd($A, B$) up to $A+B$ (this holds trivially when $A=B$).

(Sorry, I had to edit this answer several times because parts disappeared, until I realized that my inequality signs were being parsed as HTML tag starts even after dollar signs.)

Not an answer but rather a good thing to look at in connection with the problem-

http://numb3rs.wolfram.com/501/puzzle.html

Yes. The answer follows from Bezout's theorem which says that given integers A,B and C, C can be written as XA+YB if and only if C is a multiple of the highest common factor of A and B. Euclid's algorithm tells you how to compute X and Y.

It is not too hard to see that the only volumes you can get are ones of the that are integer linear combinations of A and B and you can get every positive volume that arises in this way (as long as you have a large enough additional container to store it all).