Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# Football Crazy Hockey Mad

We received a number of solutions for this problem with various degrees of explanation and completeness. Some of them are below. Well done everyone.

Adam from Bradfield School, Karina from The Manor Preparatory School and Stefan from Ecclesfield School gave a solution to the first part of the problem:

Each team wins two matches and loses two matches

Jack from Ecclesfield School offered some explanation for both parts of the question, with a similar explanation offered by Matt of Tapton School and Rebecca of The Chase High School:

The number of games possible has to be divisible by the number of teams (editor`s comment: Why?). In the first league there are 10 possible games and 5 teams, so that is 2 games they have to win each.

In the second league the number of games possible was 15 and the number of teams was 6 so that wasn`t a whole number of games.

One solution for the first league would be:

Team 1 vs Team 2 (Team 1 won)

Team 1 vs Team 3 (team 1 won)

1 vs 4 (4 won)

1 vs 5 (5 won)

2 vs 3 (2 won)

2 vs 4 (2 won)

2 vs 5 (5 won)

3 vs 4 (3 won)

3 vs 5 (3 won)

4 vs 5 (4 won)

Matt from Tapton Comprehensive sent the folowing solution:

It is possible, for each of the five Football teams must win 2 games with same score and lose 2 games with the same score as that with which they won.

The reason why you could have 5 teams is because there would be 10 matches and 5 teams and 10 divides by 5 and that equals 2 games each that each team must win.

It would be impossible because each Hockey team could either win 0 games and lose 5 games, win 1 game and lose 4 games, win 2 games and lose 3 games, win 3 games and lose 2 games, win 4 games and lose 1 game or win all 5 games. But none of these would be even. The reason why you could not have six teams leading because there are 15 matches and 6 teams and 15 will not divide into 6 exactly.

Finally, a more general solution offered by Andrei of School 205 Bucharest:

I observed that, if there were 5 teams, there would be 10 matches played. This result was obtained by adding all possible combinations:

The first team has to play with all other 4 (4 matches)

The second team also plays 4 matches, but one was already taken into account (3 matches).

The third team plays 4 matches, but two were already taken into account (2 matches).

The fourth team plays 4 matches, but 3 were already taken into account (1 match).

The fifth team plays 4 matches, but all were already taken into account (no matches).

So, the total is: $4 + 3 + 2 + 1 = 4 \times5/2 = 10$ (matches).

In this case it is possible for all teams to be league leaders, as it is possible for every team to win 2 matches, and to lose 2:

$10/5 = 2$ (won matches).

Now, if there are six teams, the total number of played matches would be:

$$5 + 4 + 3 + 2 + 1 = 5 \times6/2 = 15$$ (matches).

As 15 is not divisible by 6, it`s impossible that every team will win the same number of matches. So, it`s impossible for all the teams to be league leaders.

Now, I look more generally at the problem:

If the number of teams is even ($2k$), than the total number of matches would be: $$(2k - 1) + (2k - 2) + \dots + 1 = \frac{2k(2k - 1)}{2} = k(2k - 1)$$ This result should be divided by the number of teams: $$\frac{k(2k - 1)}{2k} = \frac{(2k - 1)}{2}$$ This number will never be divisible by 2, so, if there is an even number of teams, not all of them could be league leaders.

Now, I analyze the case of an odd number of teams $(2k + 1)$; the total number of matches is: $$2k + (2k - 1) + (2k - 2) + \dots + 1 = \frac{(2k + 1)2k}{2} = k(2k + 1)$$ This number should be divided by the number of teams $(2k+1)$: $$\frac{k(2k + 1)}{2k + 1} = k$$ So, if the number of teams is odd, then all of them could be league leaders.

Note: The total number of matches played in a round robin tournament with $p$ teams, each playing only once with all others is the number of ways of selecting a group of 2 teams from a group of $p$, when order does not matter, i.e. combinations: $$^{p}C_{2} = \frac{p!}{2!(p - 2)!} = \frac{p(p - 1)}{2}$$

## You may also like

### Adding All Nine

### Summing Consecutive Numbers

Links to the University of Cambridge website
Links to the NRICH website Home page

Nurturing young mathematicians: teacher webinars

30 April (Primary), 1 May (Secondary)

30 April (Primary), 1 May (Secondary)

Or search by topic

Age 11 to 14

Challenge Level

- Problem
- Getting Started
- Student Solutions

We received a number of solutions for this problem with various degrees of explanation and completeness. Some of them are below. Well done everyone.

Adam from Bradfield School, Karina from The Manor Preparatory School and Stefan from Ecclesfield School gave a solution to the first part of the problem:

Each team wins two matches and loses two matches

Jack from Ecclesfield School offered some explanation for both parts of the question, with a similar explanation offered by Matt of Tapton School and Rebecca of The Chase High School:

The number of games possible has to be divisible by the number of teams (editor`s comment: Why?). In the first league there are 10 possible games and 5 teams, so that is 2 games they have to win each.

In the second league the number of games possible was 15 and the number of teams was 6 so that wasn`t a whole number of games.

One solution for the first league would be:

Team 1 vs Team 2 (Team 1 won)

Team 1 vs Team 3 (team 1 won)

1 vs 4 (4 won)

1 vs 5 (5 won)

2 vs 3 (2 won)

2 vs 4 (2 won)

2 vs 5 (5 won)

3 vs 4 (3 won)

3 vs 5 (3 won)

4 vs 5 (4 won)

Matt from Tapton Comprehensive sent the folowing solution:

It is possible, for each of the five Football teams must win 2 games with same score and lose 2 games with the same score as that with which they won.

The reason why you could have 5 teams is because there would be 10 matches and 5 teams and 10 divides by 5 and that equals 2 games each that each team must win.

It would be impossible because each Hockey team could either win 0 games and lose 5 games, win 1 game and lose 4 games, win 2 games and lose 3 games, win 3 games and lose 2 games, win 4 games and lose 1 game or win all 5 games. But none of these would be even. The reason why you could not have six teams leading because there are 15 matches and 6 teams and 15 will not divide into 6 exactly.

Finally, a more general solution offered by Andrei of School 205 Bucharest:

I observed that, if there were 5 teams, there would be 10 matches played. This result was obtained by adding all possible combinations:

The first team has to play with all other 4 (4 matches)

The second team also plays 4 matches, but one was already taken into account (3 matches).

The third team plays 4 matches, but two were already taken into account (2 matches).

The fourth team plays 4 matches, but 3 were already taken into account (1 match).

The fifth team plays 4 matches, but all were already taken into account (no matches).

So, the total is: $4 + 3 + 2 + 1 = 4 \times5/2 = 10$ (matches).

In this case it is possible for all teams to be league leaders, as it is possible for every team to win 2 matches, and to lose 2:

$10/5 = 2$ (won matches).

Now, if there are six teams, the total number of played matches would be:

$$5 + 4 + 3 + 2 + 1 = 5 \times6/2 = 15$$ (matches).

As 15 is not divisible by 6, it`s impossible that every team will win the same number of matches. So, it`s impossible for all the teams to be league leaders.

Now, I look more generally at the problem:

If the number of teams is even ($2k$), than the total number of matches would be: $$(2k - 1) + (2k - 2) + \dots + 1 = \frac{2k(2k - 1)}{2} = k(2k - 1)$$ This result should be divided by the number of teams: $$\frac{k(2k - 1)}{2k} = \frac{(2k - 1)}{2}$$ This number will never be divisible by 2, so, if there is an even number of teams, not all of them could be league leaders.

Now, I analyze the case of an odd number of teams $(2k + 1)$; the total number of matches is: $$2k + (2k - 1) + (2k - 2) + \dots + 1 = \frac{(2k + 1)2k}{2} = k(2k + 1)$$ This number should be divided by the number of teams $(2k+1)$: $$\frac{k(2k + 1)}{2k + 1} = k$$ So, if the number of teams is odd, then all of them could be league leaders.

Note: The total number of matches played in a round robin tournament with $p$ teams, each playing only once with all others is the number of ways of selecting a group of 2 teams from a group of $p$, when order does not matter, i.e. combinations: $$^{p}C_{2} = \frac{p!}{2!(p - 2)!} = \frac{p(p - 1)}{2}$$

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?