There are 2002 cards on the table with numbers $1, 2, 3, \ldots, 2002$. Two players take one card each in turns. After all the cards are taken, the winner is the one whose last digit of the sum of the numbers on the taken cards is greater. Who of the players can always win, regardless of how the opponent plays, and how should they play?
Clearly, only the last digits of the numbers written on the cards affect the result. Therefore, we can consider that there are 200 cards with digits $3, \ldots, 9,0$ and 201 cards with digits 1 and 2. The first player can initially take a card with the digit 2, and then repeat the second player's moves as long as possible. When he can no longer repeat the second player's move (after the second player takes the last card with the digit 1), the first player takes any of the remaining cards, and then again repeats the second player's moves. And so on. As a result, at the end, both will have an equal number of cards of each type, except for the cards with the digits 1 and 2: the first player will have one more card with the digit 2 and one less card with the digit 1.

Since the sum $100 \cdot(0+1+\ldots+9)$ ends in zero, the sum of the first player's cards ends in 2, and the sum of the second player's cards ends in 1.

## Answer

The first.