Sum of secret numbers is 101

Question

Alice, Bob, and Charlie were given three numbers, respectively. They were said that all the three numbers were positive integers and the sum was 101.

Alice: I know we have different numbers.

Bob: Aha, I got it. I found all the numbers.

Charlie: Me, too. I know our numbers, now.

Alice: Alas, I still don't know.

What were the three numbers?

Answer 1

Alice: I know we have different numbers.

Alice has an even number larger than 50, so at least 52.

Bob: Aha, I got it. I found all the numbers.

If Bob has a number less than 47, then he can't tell if Alice's number if 52 or 54. Thus Bob's number is 47 or 48, and Alice's is 52. (Bob's number is at most 48 since Alice's is at least 52)

Charlie: Me, too. I know our numbers, now.

Well, duh. Assuming Bob is competent with logic, his number is the last one needed to fill in this gap.

Alice: Alas, I still don't know.

We don't, either. The possible choices now are $(a,b,c)=(52,48,1)$ or $(52,47,2)$

Answer 2

The answer is:

Alice has 52, Bob has 47 or 48, and Charlie has 2 or 1 (respectively)

Alice: I know we have different numbers.

So no one can have the same number as Alice, it must be 51 or over. Also, so Bob and Charlie cannot have the same number, the remainder must be odd. Namely, Alice's number is even. Therefore, Alice's number is in {52, 54, ... 98}.

Bob: Aha, I got it. I found all the numbers.

If Bob has 48, he would know that the others are 52 and 1. If he has 47, he would know that the others are 52 and 2. But if he has anything lower than that, there would always be more than one option for the others. For example, of he has 46, than Alice and Charlie can have 52 and 3 OR 54 and 1. So Bob has 47 or 48 (and Alice has 52).

Charlie: Me, too. I know our numbers, now.

Of course - if he has 1, than Bob has 48. If he has 2, Bob has 47. And Alice has 52.

Alice: Alas, I still don't know.

Indeed, she doesn't. Nor do we...

Answer 3

Here is what we know:

1. All numbers are positive and integers
2. They are different
3. Bob could deduce all 3 numbers just by knowing his number
4. Charlie could know all 3 numbers by knowing his number and knowing that Bob could deduce all three numbers by knowing his number (point 3 above)
5. Alice could not deduce her number after knowing points 1-4

So now, lets find the numbers: Bob could deduce his number just by looking at his number. Which means he had a number, which coupled with just one unique combination of 2 distinct positive integers could result in a sum total of 101.

The only number I can think of is 97 or 98. If he had 97 then he knows the other 2 numbers have to be 3 and 1: 97 + 3 + 1 = 101

and if he had 98, then he knows the other two numbers had to be 2 and 1: 98 + 2 + 1 = 101

On the contrary, if he had a 96 for example (or anything lower), there exists more than one combination of distinct integers which could add to 101: 96 + 4 + 1 = 101 and also 96 + 3 + 2 = 101. He could never know for sure what Alice and Charlie have.

Now lets look at Charlie. Charlie being the smart guy he is, immediately put 2 and 2 together (or in this case 3 numbers which add top 101). He knows Bob either has 98 or 97. To get a definite answer he looks at his own number. If it is a 2 then he knows Bob has a 98 and clueless Alice has a 1. If he has a 3, he knows Bob has a 97 and dumbfounded Alice has a 1.

Lastly, we go to Alice, she is at a disadvantage her. She has figured the set is {97,3,1} or {98,2,1} but since both result in her having a 1, she cannot pick which case is true here and hence cannot pick the correct set.

FINAL ANSWER: {97,3,1} or {98,2,1}

Note: the question asks for the 3 numbers and not who had what number. So When Bob says he knows the numbers, he knows the 3 numbers without knowing who had what. this changes when Charlie proclaims he knows the 3 numbers as well, Bob and Charlie now know the 3 numbers and who had what.