BCS Christmas Competition

[StuartR]

I really enjoyed this puzzle that came in an email from BCS yesterday. I'm pretty sure that I have solved it correctly, but it wasn't easy.

Amy, Bob and Carol sit in a circle wearing Christmas hats. Each hat has a number on it so all three can see the others numbers but not their own. They are told that the numbers are three different positive digits. They are each to make a statement in turn and are told to raise their hands when they know their own number.

Amy says: "Carol's number is greater than Bob's". No one raises their hand.
Bob then says: "The sum of Amy's and Carol's numbers is even. On hearing this, Carol raises her hand. Even then Amy and Bob do not raise theirs.
But, after a pause, and once it has become clear that Amy isn't going to raise her hand, Bob raises his and then Amy raises hers.

What are Amy's, Bob's and Carol's numbers?

The competition rules said that your submission must show the method used to determine the solution.

[John Gray]

It seems to me that the solution cannot be unique. For example, A=1, B=2 and C=3 fits - and the chap and ms-chaps can raise their hand any time they like, for they are competitionally perverse! (Think bluffing in poker...)

[StuartR]

I think you can assume that this is a logic puzzle, and that the chaps only raise their hands when they are sure that they know their own number. I also assume that the statements they make are all true.

Given these reasonable assumptions there does seem to be one correct answer.

[StuartR]

I'm surprised nobody has had a go at this. It's not easy but it only took me a few hours and I'm useless at these puzzles.

[John Gray]

So I am not even nobody, huh?!

[HansV]

Stuart,

I looked at the puzzle on and off since you posted it, but didn't solve it until now because I didn't take the time to sit down and concentrate on it.
If I'm not mistaken, the solution is

Spoiler

Amy has 6, Bob has 7 and Carol has 8

The reasoning is rather lengthy: I successively excluded possibilities, but I saved an Excel spreadsheet with the decisions I made. I can post it if desired.

[StuartR]

That's the same answer as I got, and I also had to successively eliminate possibilities

[StuartR]

So I am not even nobody, huh?!

Fairy 'nuff. You did have a go, but I'm pretty sure you were cheating!

[Skitterbug]

That's the same answer as I got, and I also had to successively eliminate possibilities

I was wondering if

Spoiler

Amy - 3, Bob - 4 and Carol - 5 would work? I had this guess before I peeked at Hans' answer. I'm definitely not a math student So I may be showing my lack of knowledge big time here.......

[StuartR]

Skitter,

Let's assume that you are correct, Amy=3, Bob=4, Carol=5.

Carol can see that Amy has 3 and Bob has 4. After hearing the first two statements she knows that she has an odd number bigger than 4. This could be 5, 7 or 9. There is no way for her to tell which of these she has, so she could not put her hand up.

[Skitterbug]

Skitter,

Let's assume that you are correct, Amy=3, Bob=4, Carol=5.

Carol can see that Amy has 3 and Bob has 4. After hearing the first two statements she knows that she has an odd number bigger than 4. This could be 5, 7 or 9. There is no way for her to tell which of these she has, so she could not put her hand up.

So far so good. I didn't check the link you provided in the starting post to this thread. Maybe I ought to do that.
Thanks for keeping my brain working even if it doesn't always come to the right conclusions!
Merry Christmas!

ps: I see what the BCS link is now.........

[Becks]

The link didn't yield a path to the solution, so my two cents worth..
Assuming zero isn't a digit and players are playing to win.

The sequence B = 7 C = 8 should be one of first eliminated, because if A = 9 then C would have the automatic win, so Amy wouldn't risk that clue. Same reason for eliminating B = 7 and C = 9 (what if A = 8), and also C != 3 for similar logic (if A were 1 or 2 then B would have automatic win).

Regards
Kevin

[StuartR]

You are on the right lines, but can eliminate whole classes of possible values fairly easily still