Given two empty jugs with capacities a and b liters and an infinite water supply, determine a way to end up with exactly X liters of water.

This is a classic problem that can be solved using the principles of the Extended Euclidean Algorithm. The core idea is that any amount of water achievable, X, must be a linear combination of the jug capacities, i.e., X = ka + lb, where k and l are integers representing the number of times water is added or removed from each jug. If X is a multiple of the greatest common divisor (GCD) of a and b, then it's possible to measure X liters. The algorithm involves a sequence of filling, emptying, and pouring water between the jugs until the target amount is reached. For example, to get 4 liters with 3L and 5L jugs:

1. Fill the 5L jug.
2. Pour from the 5L jug into the 3L jug until the 3L jug is full. (5L jug now has 2L).
3. Empty the 3L jug.
4. Pour the 2L from the 5L jug into the 3L jug. (3L jug now has 2L).
5. Fill the 5L jug.
6. Pour from the 5L jug into the 3L jug until the 3L jug is full. (The 3L jug needs 1L more, so 1L is poured from the 5L jug. The 5L jug now has 4L).

For more than two buckets, the problem becomes a generalization of the two-bucket problem. The target amount X must be a linear combination of all jug capacities (X = k1a1 + k2a2 + ... + kn*an). If X is a multiple of the GCD of all jug capacities, then it's theoretically possible. The solution would involve a more complex state-space search (like Breadth-First Search or Depth-First Search) to find the sequence of operations, as the number of possible pourings and combinations increases significantly with each additional jug.