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Question:

In a chess tournament, there are four groups, each containing an equal number of players. Each player plays

1. against every other player belonging to one's own group exactly once;
2. against each player belonging to one of the remaining three groups exactly twice;
3. against each player belonging to one of the remaining two groups exactly three times; and
4. against each player belonging to the remaining group exactly four times.

If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is . 

Explanation:

Let there be n players in each of these 4 groups.
∴ There are total 4n players.

Let the groups be called G1, G2, G3 & G4.

Now, a player from G1, say A, plays with other (n - 1) players from G1 exactly once.
Hence, he plays (n -1) games.

A will play with each of n players of G2 twice, hence total 2n games.

A will play with each of n players of G3 thrice, hence total 3n games.

A will play with each of n players of G4 four times, hence total 4n games.

∴ A will play a total of (n - 1) + 2n + 3n + 4n = 10n - 1 games.

Now, each of the 4n players will play (10n - 1) games.
∴ There will be total 4n × (10n - 1)/2 games [/2 because each game is counted twice]
⇒ There are total 2n(10n - 1) games

⇒ 2n(10n - 1) > 1000
⇒ n(10n - 1) > 500

The least possible value of n satisfying the above inequality is 8.

Hence, 8.