This post contains my proposed solution to Problem #004 - solvability of the water buckets. Please do not read this solution before making a serious attempt at the problem.
A possible solution is to consider a clever invariant that applies to the amount of water that each bucket is holding at any point in time. To make this easier, let's call \(d\) to the greatest common divisor of the \(c_i\), \(i = 1, \cdots, n\) (\(d = \gcd(c_1, \cdots, c_n)\)). Let's also say the amount of water in bucket \(i\) is \(w_i\). We will show that, regardless of the moves we make, \(w_i\) is always a multiple of \(d\) for all \(i\) (which we write \(d | w_i\) for _"\(d\) divides \(w_i\)"_).
At the start all buckets are empty, so \(w_1 = \cdots = w_n = 0\) and \(0\) is a multiple of \(d\) so that is that. Now we show that the three moves above preserve this property that \(d | w_i\ \forall i\).
- Emptying bucket \(i\): this means \(w_i = 0\) and \(d | 0\) so everything is good;
- Filling bucket \(i\): this means \(w_i = c_i\) but, by definition, \(d\) is a divisor of \(c_i\) so certainly we have \(d | c_i\);
- Moving water from bucket \(i\) to bucket \(j\) until either bucket \(i\) becomes empty or bucket \(j\) becomes full, whatever happens first: before we move water around we have that \(d | w_i\), \(d | w_j\) so we can say that \(w_i = k_i d\) and \(w_j = k_j d\) for some integer values of \(k_i, k_j\). Now when we start moving the water, we have to analyze what happens depending on whether bucket \(i\) becomes empty and \(j\) is not full yet or bucket \(j\) becomes full while \(i\) possibly has some water left:
- if bucket \(i\) becomes empty then \(w_i = 0\) and \(w_j = (k_i d) + (k_j d) = (k_i + k_j) d\); \(d | 0\) and \(d | (k_i + k_j) d\) so everything stays a multiple of \(d\);
- if bucket \(j\) got full, then \(w_j = c_j\) and \(d | c_j\), so this is ok; we just need to check if the amount of water left in bucket \(i\) is a multiple of \(d\) or not. Well, bucket \(j\) had \(k_j d\) water and now has \(c_j\), so bucket \(i\) gave \(c_j - k_j d\) water to bucket \(j\). If bucket \(i\) had \(k_i d\) water it now has \(w_i = k_i d - (c_j - k_j d)\). But this is still a multiple of \(d\) because \(c_j\) was! We can write \(c_j = k d\) with \(k\) integer, showing that \(w_i = k_i d - (c_j - k_j d) = d(k_i - k + k_j)\) is a multiple of \(d\)!
We showed that no matter what we do, the amount of water in a bucket is always a multiple of \(d\), so if \(t\) is not a multiple of \(d\) this means we can never have a single bucket holding \(t\) litres of water...
If you have any questions about my solution, found any mistakes (whoops!) or would like to share your solution, be sure to leave a comment below.