Suman Saraf rightly reminded me that I had not posted the solution for this puzzle. My apologies. Many people did this correctly. Mahi Saraf, Anisha Sharma Goyal, Pratik Poddar, Abhinav Jain and Vishal Poddar all sent correct answers – well done!

Answer is Charlie. Here is the explanation:

First, realize that the total points awarded is 40. Given that we are dealing with integers that means the total number of Events (TE) times the total number of points awarded in an event (TP) must equal 40. We can quickly eliminate some possibilities.

Events (TE) |
Points (TP) |

1 |
40 |
Not possible since we know there are at least two events |

2 |
20 |
Not possible since Adam and Bob both won 1 event. There would be no way for Bob and Charlie to finish with the same points. |

4 |
10 |
A possibility |

5 |
8 |
A possibility |

8 |
5 |
Not possible, since 1st>2nd>3rd>0 means the minimum number of points awarded HAS to be 6 (3, 2, 1) |

10 |
4 |
Same Reason |

20 |
2 |
Same Reason |

40 |
1 |
Same Reason |

So we now know it was either 4 events with 10 points awarded in an event or 5 events with 8 points awarded. Let’s look at the 4/10 situation first. The possible points for each place with 10 overall points available is (all other combinations are impossible because of the 1st>2nd>3rd>0 constraint):

- 5, 3, 2Not possible as there is no way to get to 22 points in 4 events (greatest possible points would be 20)
- 5, 4, 1Same reason
- 6, 3, 1Not possible, since no combination of 4 numbers can get to 22 points (3 ×6+3= 21, 4 ×6=24)
- 7, 2, 1A possibility – let’s look further.

Can we get Adam to 22? Yes. Adam can finish 1st 3 times and 3rd once (in the Javelin). 3 ×7+1 = 22.

Can we get Bob to 9? Nope. Bob won the Javelin (7 points). There’s no way to get to 9 with the remaining event/point combinations. Sooooo …

There must be 5 events with a total of eight points (we’re getting close). Let’s look at the possible points (again, the other combinations are not possible because of the 1st>2nd>3rd>0 constraint):

- 4, 3, 1Not possible as there is no way to get to 22 points (4 ×5=20 is maximum)
- 5, 2, 1Looks like this is the winner. Let’s check.

Can we get Adam to 22? Yes. Adam can finish 1st 4 times and 2nd once (4 ×5+2). Since we know Bob finished 1st in the Javelin, Adam must have finished 2nd.

Can we get Bob to 9? Yes. Bob finished 1st in the Javelin (5 points) and finished 3rd 4 times (4 ×1) for a total of 9 points.

Can we get Charlie to 9? Yes. Charlie finished 3rd in the Javelin (1 point) and must have finished 2nd in the 4 other events (4 ×2) for a total of 9 points.

Since Charlie finished 2nd in EVERY EVENT OTHER THAN THE JAVELIN, he MUST have finished 2nd in the 100-meter dash. Here is a little table:

Javelin |
Event 2 |
Event 3 |
Event 4 |
100m Dash |
Total |

Adam |
2nd (2 pts) |
1st (5 pts) |
1st (5 pts) |
1st (5 pts) |
1st (5 pts) |
22 pts |

Bob |
1st (5 pts) |
3rd (1 pt) |
3rd (1 pt) |
3rd (1 pt) |
3rd (1 pt) |
9 pts |

Charlie |
3rd (1 pt) |
2nd (2 pts) |
2nd (2 pts) |
2nd (2 pts) |
2nd (2 pts) |
9 pts |