, 2015

Answer: $\frac{5}{2}K$.
Each week, the accumulation of laundry increases the total amount by $K$, after which the washing decreases it by at least one third, because, by the pigeon-hole principle, the bin with the most laundry must contain at least a third of the total. Hence the amount of laundry post-wash after the $n$th week is bounded above by the sequence $a_{n+1} = \frac{2}{3}(a_n + K)$ with $a_0 = 0$, which is clearly bounded above by $2K$. The total amount of laundry is less than $2K$ post-wash and $3K$ pre-wash.

Now suppose pre-wash state $(a, b, c)$ precedes post-wash state $(a, b, 0)$, which precedes pre-wash state $(a', b', c')$. The relations $a \le c$ and $a' \le a + K$ lead to
$$3K > a + b + c \ge 2a \ge 2(a' - K),$$
and similarly for $b'$, whence $a', b' < \frac{5}{2}K$. Since also $c' \le K$, a pre-wash bin, and a fortiori a post-wash bin, always contains less than $\frac{5}{2}K$.

Consider now the following scenario. For a start, we keep packing the three bins equally full before washing. Initialising at $(0, 0, 0)$, the first week will end at $(\frac{1}{3}K, \frac{1}{3}K, \frac{1}{3}K)$ pre-wash and $(\frac{1}{3}K, \frac{1}{3}K, 0)$ post-wash, the second week at $(\frac{5}{9}K, \frac{5}{9}K, \frac{5}{9}K)$ pre-wash and $(\frac{5}{9}K, \frac{5}{9}K, 0)$ post-wash, &c. Following this scheme, we can get arbitrarily close to the state $(K, K, 0)$ after washing. Supposing this accomplished, placing $\frac{1}{2}K$ kg of laundry in each of the non-empty bins leaves us in a state close to $(\frac{3}{2}K, \frac{3}{2}K, 0)$ pre-wash and $(\frac{3}{2}K, 0, 0)$ post-wash. Finally, the next week's worth of laundry is directed solely to the single non-empty bin. It may thus contain any amount of laundry below $\frac{5}{2}K$ kg. $\square$