20
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ all positive integer numbers? $\endgroup$ – Oray Aug 17 '17 at 9:12
  • $\begingroup$ @Oray Oh, I missed. Yes, they were all positive. $\endgroup$ – P.-S. Park Aug 17 '17 at 9:12
  • $\begingroup$ Very close to this puzzle $\endgroup$ – Rob Watts Aug 17 '17 at 17:03
16
$\begingroup$

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 $(52,47,2)$

$\endgroup$
  • 1
    $\begingroup$ Ooooops! Sorry. My mistake. Could you read again after I edited it? $\endgroup$ – P.-S. Park Aug 17 '17 at 9:29
  • 1
    $\begingroup$ Alice's sentence should be corrected to "we have different numbers." $\endgroup$ – P.-S. Park Aug 17 '17 at 9:31
  • 2
    $\begingroup$ Then Alice can work out the three numbers! Actually, Alice has at least as much information as us, so Alice can always work out the three numbers, if we are able to. $\endgroup$ – Wen1now Aug 17 '17 at 9:32
  • 2
    $\begingroup$ Right! It's better to correct the numbers in Bob's deduction. $\endgroup$ – P.-S. Park Aug 17 '17 at 9:44
  • $\begingroup$ Bob can have 47 or 48, as shown in the final answer, not 48 or 49 as stated in Bob's deduction. $\endgroup$ – saulspatz Aug 18 '17 at 2:54
12
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ I don't get it. Why Alice must have 51 or over? Why not 49? Then 3 numbers would be 49, 50, 2. What's wrong in there? $\endgroup$ – jack Aug 18 '17 at 18:46
  • $\begingroup$ @mikele Had she had 49, the 3 numbers could also have been (49, 49, 3). This is true for any number smaller than 51 - someone else could have had the same number as her. The fact that she knows that all numbers are different tells us that her number is 51 or larger. $\endgroup$ – Angkor Aug 19 '17 at 19:46
  • $\begingroup$ But it could have been also (49, 47, 5), so all 3 different numbers. I don't know. Personally the questions doesn't satisfy me. $\endgroup$ – jack Aug 19 '17 at 21:43
  • $\begingroup$ @mikele Alice knows for certain that no number is the same as hers. If there is any possibility of a=b or a=c, then she cannot state that no one has the same number as hers. Since (49, 49, 3) is possible, then she would not have made the statement she made if her number were 49. $\endgroup$ – LeppyR64 Aug 21 '17 at 11:59
-1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ If Alice is holding 1, 2, or 3, how does she know for certain that the other two numbers are unique? If instead you are implying that she is holding 97 or 98, then if Bob is holding 1, 2, or 3 how does he know what all three numbers are knowing only what number he is holding (1,2, or 3) and that Alice has told them all three numbers are unique? $\endgroup$ – LeppyR64 Aug 21 '17 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.