310. There are 60 coins on the table. Two players take turns picking up $1, 2, 3$ or 4 coins from the table. The player who picks up the last coins wins. Who will win with the correct play?
$\triangle$ Let's consider several cases, starting with a small number of coins on the table.

1) Suppose there are 1, 2, 3, or 4 coins on the table initially. Then the first player wins by taking all these coins on the first move.
2) Suppose there are 5 coins. Here, the second player wins by complementing to 5: if the first player takes 1, 2, 3, or 4 coins on the first move, the second player should take 4, 3, 2, or 1 coins in response.
3) Suppose there are 6, 7, 8, or 9 coins on the table initially. Then the first player wins by taking 1, 2, 3, or 4 coins on the first move, reducing this case to the previous one but with the players' roles reversed.
4) If there are 10 coins on the table, the second player wins by complementing the number of coins taken by the first player on the same move to 5, twice.

In general, the following pattern emerges: if the number of coins initially on the table is divisible by 5, the second player wins; if it is not divisible by 5, the first player wins. In this case, 60 is divisible by 5.

Answer: the second player.