## Task A-4.8. (10 points)

On the table, there are 1234 pebbles. Ratko and Rudi are playing the following game: first, Ratko takes an even number of pebbles, at least two, but no more than 100, and then Rudi takes an odd number of pebbles, at least one, but no more than 99. The moves continue alternately, adhering to the same conditions. The player wins if they pick up all the pebbles or if the other player cannot make their move. Who has the winning strategy, i.e., can win regardless of the other player's moves?
## First Solution.

Notice that each player can achieve taking 101 stones in two consecutive moves. For example, Ratko can ensure that he and Rudi together take 101 stones in their two consecutive moves. If Rudi takes 1 stone, Ratko takes 100 stones; if Rudi takes 3 stones, Ratko takes 98, and so on.

Since $1234 = 12 \cdot 101 + 22$, Ratko should play as follows:

In his first move, Ratko takes 22 stones.

After that, if Rudi takes $x$ stones, Ratko takes $101 - x$ stones (which is an even number, since $x$ is odd).

After Ratko's twelfth move, there will be 101 stones left on the table.

After Rudi's move, there will be an even number (at most 100) of stones left on the table, which Ratko can take all and thus win.

4 points

Therefore, Ratko has a winning strategy.

Note: If a student writes only that Ratko has a winning strategy without justification, they receive 0 points.