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There are three piles of rocks: in the first pile there are 10 rocks, 15 in the second pile and 20 in the third pile. In this game (with two players), in one turn a player is allowed to divide one of the piles into two smaller piles. The loser is the one who cannot make a move. Which player would be the winner?

a$)$ There is an unlimited set of cards with the words “abc”, “bca”, “cab” written. Of these, the word written is determined according to this rule. For the initial word, any card can be selected, and then on each turn to the existing word, you can either add on a card to the left or to the right, or cut the word anywhere $($between the letters$)$ and put a card there. Is it possible to make a palindrome with this method?

b$)$ There is an unlimited set of red cards with the words “abc”, “bca”, “cab” and blue cards with the words “cba”, “acb”, “bac”. Using them, according to the same rules, a palindrome was made. Is it true that the same number of red and blue cards were used?

At a contest named “Ah well, monsters!”, 15 dragons stand in a row. Between neighbouring dragons the number of heads differs by 1. If the dragon has more heads than both of his two neighbors, he is considered cunning, if he has less than both of his neighbors – strong, the rest $($ including those standing at the edges $)$ are considered ordinary. In the row there are exactly four cunning dragons – with 4, 6, 7 and 7 heads and exactly three strong ones – with 3, 3 and 6 heads. The first and last dragons have the same number of heads.

a$)$ Give an example of how this could occur.

b$)$ Prove that the number of heads of the first dragon in all potential examples is the same.

a$)$ There are 21 coins on a table with the tails side facing upwards. In one operation, you are allowed to turn over any 20 coins. Is it possible to achieve the arrangement were all coins are facing with the heads side upwards in a few operations?

b$)$ The same question, if there are 20 coins, but you are allowed to turn over 19.

We are given a table of size $n \times n$. $n-1$ of the cells in the table contain the number $1$. The remainder contain the number $0$. We are allowed to carry out the following operation on the table:

1. Pick a cell.

2. Subtract 1 from the number in that cell.

3. Add 1 to every other cell in the same row or column as the chosen cell.

Is it possible, using only this operation, to create a table in which all the cells contain the same number?