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Wow, we received lots of solutions to this problem! It offered a good opportunity to think in a logical, systematic way.

Lots of people got the answer to Medals Count right. Rachel, at Milton Keynes Academy, gave us a great explanation of her method:

There are 38 bronze medals in total for the 3 countries. France has 18 of them, so there are 20 bronze medals between Italy and Japan. These are the possible combinations, needing to be over 6:

Italy: either 6, 7, 8, 9, 10, 11, 12 , 13 or 14 bronze

Japan: 14, 13, 12, 11, 10, 9, 8, 7 or 6 bronze

Italy must have at least 8 gold, because they have more than France (who have 7). So, they need to have at least 10 bronzes as they have 2 more bronzes than golds. This means that Italy has at most 9 silvers.

Japan has at most 10 bronzes because there are 20 bronzes between Japan and Italy. They have one more gold than Italy, giving Japan at least 9 golds. Japan also have 3 less silvers than Italy, giving them at most 6 silvers. France has twice as many silvers as Italy has golds (8x2 = 16), so France has at least 16 silvers.

Now let's explore the possibilities for Italy's bronze medals. Italy can have 10, 11, 12, 13 or 14 bronzes. But if Italy has 11 or more bronzes, then it has 9 or more golds, and so it has at most 7 silvers. This means that Japan has at most 4 silvers, which isn't allowed. So Italy must have 10 bronzes exactly.

In conclusion:

Italy: 10 bronzes 9 silvers 8 golds

Japan: 10 bronzes 6 silvers 9 golds

France: 18 bronzes 16 silvers 7 golds

Excellent! Luke, from Cottenham Village College, did things more algebraically:

First, give the number of medals names:

Now apply the rules given:

From rule 1: g=d-1, h=e+3

From rule 2: c=18, a=7

From rule 5: i=d+1

From rule 7: b=2d-2

The table now looks like this:

Apply rule 6: f+d+1+18=38, and the table looks like:

As France has the fewest golds, d-1 must be at least 8 so d is at least 9.

As 2d+e+3 = 27 (rule 4), and e must be at least 6 (Japan must have at least 6 silver medals: rule 3), we see that d = 9.

This makes e = 6.

The table finally looks like this:

Great - thanks! How about Football Champ? The Year 9 class at Ounsdale High School showed us their method:

And their results were as follows:

Nicely done!

For the Fencing Tournament problem, Szymon from Earl Mortimer College recommends making a table. He explains his method as follows:

Tip 2 says 'Daphne won her match against Becky'. Mark this in the table as follows:

Thanks! Finally, the Hockey problem - this was tough! A bit of careful reading tells you that the results given are only a**partial** list of results - **some games may not have been played yet**. Samantha, from Laxey Primary School, gives her answer as follows:

and the individual results:

A v B: B wins 2-0

A v C: draw 2-2

A v D: A wins 2-0

B v C: draw 2-2

B v D: B wins 1-0

C v D: not played yet.

Thanks to everyone who sent in solutions this month!

Lots of people got the answer to Medals Count right. Rachel, at Milton Keynes Academy, gave us a great explanation of her method:

There are 38 bronze medals in total for the 3 countries. France has 18 of them, so there are 20 bronze medals between Italy and Japan. These are the possible combinations, needing to be over 6:

Italy: either 6, 7, 8, 9, 10, 11, 12 , 13 or 14 bronze

Japan: 14, 13, 12, 11, 10, 9, 8, 7 or 6 bronze

Italy must have at least 8 gold, because they have more than France (who have 7). So, they need to have at least 10 bronzes as they have 2 more bronzes than golds. This means that Italy has at most 9 silvers.

Japan has at most 10 bronzes because there are 20 bronzes between Japan and Italy. They have one more gold than Italy, giving Japan at least 9 golds. Japan also have 3 less silvers than Italy, giving them at most 6 silvers. France has twice as many silvers as Italy has golds (8x2 = 16), so France has at least 16 silvers.

Now let's explore the possibilities for Italy's bronze medals. Italy can have 10, 11, 12, 13 or 14 bronzes. But if Italy has 11 or more bronzes, then it has 9 or more golds, and so it has at most 7 silvers. This means that Japan has at most 4 silvers, which isn't allowed. So Italy must have 10 bronzes exactly.

In conclusion:

Italy: 10 bronzes 9 silvers 8 golds

Japan: 10 bronzes 6 silvers 9 golds

France: 18 bronzes 16 silvers 7 golds

Excellent! Luke, from Cottenham Village College, did things more algebraically:

First, give the number of medals names:

gold | silver | bronze | |

France | a | b | c |

Japan | d | e | f |

Italy | g | h | i |

Now apply the rules given:

From rule 1: g=d-1, h=e+3

From rule 2: c=18, a=7

From rule 5: i=d+1

From rule 7: b=2d-2

The table now looks like this:

gold | silver | bronze | |

France | 7 | 2d-2 | 18 |

Japan | d | e | f |

Italy | d-1 | e+3 | d+1 |

Apply rule 6: f+d+1+18=38, and the table looks like:

gold | silver | bronze | |

France | 7 | 2d-2 | 18 |

Japan | d | e | 19-d |

Italy | d-1 | e+3 | d+1 |

As France has the fewest golds, d-1 must be at least 8 so d is at least 9.

As 2d+e+3 = 27 (rule 4), and e must be at least 6 (Japan must have at least 6 silver medals: rule 3), we see that d = 9.

This makes e = 6.

The table finally looks like this:

gold | silver | bronze | |

France | 7 | 16 | 18 |

Japan | 9 | 6 | 10 |

Italy | 8 | 9 | 10 |

Great - thanks! How about Football Champ? The Year 9 class at Ounsdale High School showed us their method:

- Josh said that to work out the wins, draws and losses we looked at the points and the only possibilities are given below.
- B and C must have drawn because they each had 1 draw.
- A must have beaten B, and C must have beaten A.
- Jordan said: the maximum that team B could have scored in their draw was 2, so it was either 2 - 2 or 1- 1.
- As C only let in 2 goals in total, and C must have scored at least one goal against A, the same between B and C must have ended up 1-1.
- Total number of goals for and against must be equal, because if A scored against B it would count for A and against B, for example.

And their results were as follows:

Teams | Games Played | Won | Drawn | Lost | Goals For | Goals Against | Points |
---|---|---|---|---|---|---|---|

A | 2 | 1 | 0 | 1 | 5 | 3 | 3 |

B | 2 | 0 | 1 | 1 | 2 | 5 | 1 |

C | 2 | 1 | 1 | 0 | 3 | 2 | 4 |

Nicely done!

For the Fencing Tournament problem, Szymon from Earl Mortimer College recommends making a table. He explains his method as follows:

Tip 2 says 'Daphne won her match against Becky'. Mark this in the table as follows:

A | B | C | D | E | F | |

A | ||||||

B | ||||||

C | ||||||

D | ||||||

E | ||||||

F |

- Tip 3: 'Alice and Elsie won the same, odd, number of matches, but Alice lost to Elsie'. You know that Elsie won against Alice, so you can plot these straight away. Now, they won either 1, 3 or 5 games each, but it can't be five, because only one person could have won all their matches.
- Tip 5 says Charlotte won only one match, against the other person who won only one as well. We don't know who this is, but there is only one of them, so it can't be Alice and Elsie, so they must have won 3 of their matches...

A | B | C | D | E | F | |

A | ||||||

B | ||||||

C | ||||||

D | ||||||

E | ||||||

F |

Thanks! Finally, the Hockey problem - this was tough! A bit of careful reading tells you that the results given are only a

Team | Played | Won | Drawn | Lost | For | Against | Points |

A | 3 | 1 | 1 | 1 | 4 | 4 | 3 |

B | 3 | 2 | 1 | 0 | 5 | 2 | 5 |

C | 2 | 0 | 2 | 0 | 4 | 4 | 2 |

D | 2 | 0 | 0 | 2 | 0 | 3 | 0 |

and the individual results:

A v B: B wins 2-0

A v C: draw 2-2

A v D: A wins 2-0

B v C: draw 2-2

B v D: B wins 1-0

C v D: not played yet.

Thanks to everyone who sent in solutions this month!