Daily Math Puzzle: 2026-07-30
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2026-07-30
In a 'Championship Cup' tournament, every team played every other team exactly once. A win earns 3 points, a draw earns 1 point, and a loss earns 0 points. At the end of the tournament, the sum of all points accumulated by all teams was 51. How many teams participated in the tournament?
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Solution
7 teams — Let N be the number of teams participating in the tournament.
Since every team played every other team exactly once, the total number of matches (M) played in the tournament is given by the formula M = N * (N - 1) / 2.
Let W be the number of matches that resulted in a win/loss (one team gets 3 points, the other 0).
Let D be the number of matches that resulted in a draw (both teams get 1 point).
The total points awarded in a win/loss match are 3 (3+0), and in a draw match are 2 (1+1).
The sum of all points accumulated by all teams is given as 51. So, we have the equation: 3W + 2D = 51.
We also know that the total number of matches is M = W + D. We can express D as D = M - W.
Substitute this into the points equation:
3W + 2(M - W) = 51
3W + 2M - 2W = 51
W + 2M = 51
Now, substitute M = N * (N - 1) / 2 into the equation:
W + 2 * [N * (N - 1) / 2] = 51
W + N * (N - 1) = 51
From this, we can express W as: W = 51 - N * (N - 1).
Since W represents the number of matches with a winner, it must be a non-negative integer (W ≥ 0).
So, 51 - N * (N - 1) ≥ 0
N * (N - 1) ≤ 51 (Inequality 1)
Also, the number of wins (W) cannot exceed the total number of matches (M).
So, W ≤ M
Substitute the expressions for W and M:
51 - N * (N - 1) ≤ N * (N - 1) / 2
To solve for N, move N * (N - 1) terms to one side:
51 ≤ N * (N - 1) + N * (N - 1) / 2
51 ≤ (3/2) * N * (N - 1)
Multiply both sides by 2/3:
51 * (2/3) ≤ N * (N - 1)
34 ≤ N * (N - 1) (Inequality 2)
Now we need to find an integer N (number of teams) that satisfies both inequalities:
34 ≤ N * (N - 1) ≤ 51
Let's test integer values for N:
- For N = 1 to 5, N*(N-1) is less than 34.
(e.g., N=5, N*(N-1) = 5*4 = 20)
- If N = 6, N*(N-1) = 6 * 5 = 30. This does not satisfy N*(N-1) ≥ 34.
- If N = 7, N*(N-1) = 7 * 6 = 42.
- This satisfies 34 ≤ 42 (True) and 42 ≤ 51 (True).
- So, N=7 is a possible number of teams.
- For N=7: M = 7 * 6 / 2 = 21 matches. W = 51 - 42 = 9 wins. D = 21 - 9 = 12 draws.
- Check total points: (3 * 9) + (2 * 12) = 27 + 24 = 51. This is consistent.
- If N = 8, N*(N-1) = 8 * 7 = 56.
- This satisfies 34 ≤ 56 (True), but does not satisfy 56 ≤ 51 (False, as 56 > 51).
- So, N=8 is not a possible number of teams.
- For any N > 8, N*(N-1) will be even greater than 56 and thus will also not satisfy N*(N-1) ≤ 51.
The only value of N that satisfies all conditions is 7.
Therefore, 7 teams participated in the tournament.
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