Shaovalov A.V.

a) There are three identical large vessels. One contains 3 liters of syrup, another contains 20 liters of water, and the third is empty. You can pour all the liquid from one vessel into another or into the sink. You can choose two vessels and pour from the third into one of them until the levels of liquid in the chosen vessels are equal. How can you obtain 10 liters of 30% diluted syrup?

b) The same, but with \( N \) liters of water. For which integer values of \( N \) can you obtain 10 liters of 30% diluted syrup?
Let's denote the vessels with syrup, water, and the empty one as $A, B$, and $C$ respectively.

a) Pour 3 liters of water from $B$ to $C$, and then pour these 3 liters into the sink; repeat this operation several times until 5 liters of water remain in $B$, and in the last operation, pour the water from $C$ not into the sink, but into $A$. Now, $A$ contains 6 liters of the mixture, and $C$ is empty again. Pour 5 liters from $A$ to $C$. Now, 4 liters of water can be poured from $B$ to $A$, and the remaining liter in $B$ can be poured into the sink. Finally, return the 5 liters of the mixture from $C$ to $A$, resulting in 10 liters of the mixture: 3 liters of syrup and 7 liters of water, which is a 30% syrup solution.

b) The algorithm described in a) works for all $N \equiv 2(\bmod 3)$. If $N \equiv 1(\bmod 3)$, the algorithm is even simpler: using $C$, pour 3 liters from $B$ into the sink until 7 liters remain in $B$; then pour these 7 liters into $A$.

If $N$ is a multiple of 3, the volume of liquid in each vessel is initially a multiple of 3. The allowed operations cannot change this property, so it is impossible to obtain 10 liters of liquid (regardless of its composition) in one vessel.

## Answer

b) For any $N$ not a multiple of 3 and greater than 6.