### Tuesday Puzzle

Here’s a puzzle I hadn’t seen before. I’m concealing the source to discourage Googling, but will give credit where it’s due in a couple of days.

Alice, Betty and Carol each has a positive integer stamped on her forehead. They know that one of their numbers is equal to the sum of the other two. They proceed alphabetically around the table, each one either announcing her own number (if she’s managed to figure it out) or announcing that she doesn’t know it.

The game proceeds as follows:

Alice: “I don’t know my number”.

Betty: “I don’t know my number”.

Carol: “I don’t know my number”.

Alice: “My number is 25″.

What are the other two numbers?

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#### 25 Responses to “Tuesday Puzzle”

1. 1 1 Brandon Berg

The two lesser integers aren’t necessarily unique?

2. 2 2 JohnW

A=25, B=10, C=15 works (I did not try to prove it to be unique)

1A) A = 5 or 25

1B) B = 10 or 40

1C) C = 15 or 35

2A) if A = 5, then C would have known either 5 or 15, but if C = A = 5,
then B would have known 10, but B did not know 10, so C must have known C != 5 so C would have known 15, but C did not know 15, therefore A != 5, therefore A = 25

3. 3 3 Mike H

Assuming the numbers don’t have to be all different… my (probably flawed) logic so far…

Everyone sees two numbers, and knows their own number is the sum or difference of the ones they can see. We know that A=25, so Betty and Carol know that their own numbers are not the same – if they were, A would be even.

The game proceeds as follows:

Alice: “I don’t know my number”.

The only way for Alice to know her number is if the numbers she sees are the same, so their difference is zero. Now everyone knows that B is not C. Note that this gives no information to anyone except us – Alice could see this plainly, Betty and Carol could figure it out fromt he fact that A is odd.

Betty: “I don’t know my number”.

Betty didn’t get any information from Alice’s answer. So, the fact that she doesn’t know her number just tells Alice and Carol that their numbers are different.

Carol: “I don’t know my number”.

Carol now knows her number is different from both Alice’s and Betty’s (and that Alice’s is 25). She already knew it was different from Betty’s, since Alice’s number is odd. If Betty’s number was 50, she would now know her own number to be 75 (since Betty’s answer reveals that it can’t be 25). Now *we* know that Betty’s number is not 50. Alice already knew that, so she has no additional information from Carol’s answer except that her number is different from Betty’s.

Alice: “My number is 25″.

Suddenly, Alice knows the number. All Alice has discovered is that her number is different from Betty’s and Carol’s, yet this is enough for her to determine her own number. Whatever B and C are, therefore, their sum or difference must equal either B or C – that is, one of them is double the other…. but, there’s no way for this to be the case, and for Alice’s number to be 25, and for B and C to both be different from A.

So, I’m stumped. Unless Alice is working in octal?

No doubt, I’ve missed something important.

4. 4 4 Bennett Haselton

@Mike the error is that when Betty says “I don’t know my number”, that reveals more than just that Alice and Carol’s numbers are different.

Betty already knows B != C. Now suppose that A = 2C. Then Betty would reason as follows: “My number is either A – C = 2C – C = C, or it’s A + 2C. But I know my number is not C, therefore my number is A + 2C.”

Therefore hen Betty says “I don’t know”, she also reveals to the group that A != 2C.

Complete proof to follow…

5. 5 5 Bennett Haselton

Here’s the proof that JohnW’s solution is unique. (Interestingly, the uniqueness proof would not necessarily be valid if A had a number other than 25.)

Each woman knows her number is either the sum or positive difference of the other two numbers. At each stage, a woman has a set of known inequalities involving her own number. (She doesn’t care about the inequalities involving only the other two numbers, since she can see those two numbers anyway.) If she were able to apply one of those inequalities to her own number, to prove that it’s either *not* the sum or *not* the positive difference of the other two, then she would only be left with the other possibility, and would know her own number. So when she says “I don’t know”, we know that eliminates any relationship between the other two numbers, which, together with a known inequality, would rule out her own number as either the sum or positive difference of the other two.

When A says “I don’t know”, we know B!=C. Because if B=C, then A knows she’s either 0 or 2B, and she can’t be 0.

Now it’s B’s turn. We have 1 known inequality, B!=C. Using this inequality, what would enable B to rule out being the sum of A and C? If A were 0, then by using B!=C, B could rule out being the sum of A and C — but this is irrelevant since A!=0 anyway. But what would enable B to rule out being the positive difference of A and C? If it were the case that A=2C, then B would know that B!=C rules out being the positive difference, so B would know she was the sum of the other two. So when B says “I don’t know”, everyone now knows A!=2C. (As well as A!=C.)

When it’s C’s turn, she knows A!=C, B!=C and A!=2C. Applying each of these in turn, and what it implies if it does not enable C to rule out being the (a) sum or (b) positive difference:
A!=C — sum — B=0 — eliminated anyway
A!=C — positive difference — A!=1/2B
B!=C — sum — A=0 — eliminated anyway
B!=C — positive difference — A!=2B
1/2A!=C — sum — eliminated anyway since 1/2A cannot be the sum of A and B
1/2A!=C — positive difference — then A!=2B (we already knew that) and A!=2/3B
And as always, when C says “I don’t know” it means A!=B.

So now it’s A’s turn again. A knows:
A!=1/2B, A!=2/3B, A!=B, A!=2B, A!=C, A!=2C

A knows she’s either the sum or positive difference of the other two. Apparently at this stage, A is able to eliminate one of those possibilities and settle on the other one (25). So either the sum or positive difference of B and C must be equal to 1/2B, 2/3B, B, 2B, C, or 2C, so that if A knows she’s not that one, she must be the other one.

Trivially with B and C positive and B!=C, the sum of B and C can’t be 1/2B, 2/3B, B, 2B, C, or 2C. So it must be the positive difference which is equal to one of those (and A eliminates that one to realize that A is the sum of B and C).

If B were the positive difference, we’d have B=C-B, so A=B+C=3B, not possible since A=25. Similarly C can’t be the positive difference.

If 2B were the positive difference, we’d have 2B=C-B, so A=B+C=4B, not possible since A=25. Similarly 2C can’t be the positive difference.

If 2/3B were the positive difference, we’d have either (i) 2/3B=B-C meaning A=B+C=4C, not possible since A=25, or (ii) 2/3B=C-B meaning A=B+C=8/3B, not possible since A=25.

So if 1/2B is the positive difference, we have either (i) 1/2B=B-C meaning A=B+C = 3/2B, not possible since A=25, or (ii) 1/2B=C-B, meaning A=5/2B — which is finally satisfiable using A=25, B=10 and C=15. Thus the solution is unique.

6. 6 6 nobody.really

Oh, I’m sorry; I was filming Alice as part of my Sleepnumber Bed commercial. Did you think she was responding to Betty and Carol?

7. 7 7 ECSL

@Bennett:
Thanks for working all of this out!

The reason A can conclude that her number is 25 is because B=10 and C=15 (although it would work for B=15 and C=10).

A can work through the implications of being the sum or difference of 10 and 15.

If the difference, then A=5, and C would have known C=15 from the fact that B told her C!=A (and she could see B=10), leaving only that possibility. But since C didn’t know her number, this is impossible.

That only leaves A=B+C=25.

8. 8 8 Harold

Congratulations to Bennett Haselton and to Alice – particularly Alice who worked all that out in her head (at least as the game was played out in my head).

9. 9 9 JohnW

ECSL:

Nope, it does not work for A=25,B=15,C=10 (at least, not on Alice’s second turn as the question stated).

10. 10 10 Al V.

So, this would work for A=5, B=2, C=3 as well.

11. 11 11 Bennett Haselton

Yeah it works for any number divisible by 5 and not divisible by 4 or 3.

12. 12 12 ECSL

@JohnW: you’re quite right, the order makes a difference. Thanks!

13. 13 13 David Wallin

I propose 10 and 15. A sees 10 and 15 and reasons she must be 5 or 25, but doesn’t know which in round one. B has (say) 10, sees 25 and 15, and reasons she must be 10 or 40, but doesn’t know which (and the fact A didn’t know yet doesn’t help her). C has 15, sees 25 and 10, and reasons she must be 15 or 35, but doesn’t know which (and the fact A and B didn’t know doesn’t help her).
A realizes that, had she been 5, B, who would have seen 5 and 15, would not know if she were 10 or 20, and would have answered the same as she did in the first round. However, C would have then had to have seen 5 and 10. C options would have been 5 and 15. But if C were 5, she would realize that B would have seen 5 and 5, and known the only option for B is 10 (can’t be 0). Thus, the fact that B did not know what she was tells C that she isn’t 5 and she would have declared 15. But she didn’t so, the starting point of this paragraph—A had been 5—can’t be true.
Thus A is 25 and declares it at the start of round two.

14. 14 14 Tim

This puzzle is incorrect, women can’t do math

15. 15 15 Mike H

@Bennett Thanks!

16. 16 16 Mike

Here is a method for solving any question of this form (regardless of what the number is and what turn the first person speaks up).

If the first person answers you know the numbers are in the form:

2X, X, X

If the 2nd person answers you know the numbers are of the form:

X, 2X, X or 2X, 3X, X

If the 3rd person answers you know the numbers are of the form:

X, X, 2X or 2X, X, 3X or 2X, 3X, 5X or X, 2X, 3X

If the 4th person answers you know the numbers are of the form:

4X, 3X, X or 3x, 2X, X or 4X, X, 3X or 8X, 3X, 5X or 5X, 2X, 3X or 3X, X, 2X

To continue for the (N+1)th person, take the forms from all previous iterations, cover up the (N+1)mod3 term, and replace it with the sum of the other two. Note the difference will already have been “taken”. Remove any form that has been used already.

Note that in the given case 25 only has one unique factor, 5, so we know it has to be the form: 5X, 2X, 3X.

If person 4 (AKA 1) had said their number was 9, we would be down to 3X, X, 2X or 3X, 2X, X. In this case we wouldn’t know whether the other numbers were 6, 3 or 3, 6.

If person 4 had said their number was 8, it would be down to 4X, 3X, X or 4X, X, 3X or 8X, 3X, 5X. That is, we could not know whether the other numbers were 6, 2 or 2, 6 or 3, 5.

In other words, if the Nth person answers the Nmod3 term of the correct set must be a factor of the answer. If two forms both satisfy this condition there will be ambiguity.

17. 17 17 T.Ono

Any combination of the two numbers (B and C) that satisfies the conditions below would suffice:
1) Either B < C < 2*B or C < B < 2*B
2) The difference between the two number is 25.

I don't believe that A has to be 25. For any positive number X, we can assign it to A and still come up w/ combination of B and C that don't break the conditions above.

18. 18 18 Bill

I’ll admit that I was stumped by this problem, and have great respect for those who figured it out. However, the setup of the problem left out a key assumption.

The only way the scenario works is if each woman can trust that everyone else optimized all available information in a manner that I could not. If any woman said “I don’t know” when she could conceivably have known, her response would have provided false information to those who followed.

This is one reason why microeconomic models assuming full information and perfect rationality often predict badly. Most of us just aren’t that smart!

19. 19 19 Steve Holmes
20. 20 20 Abrey

Solving the problem from Alice’s point of view is not solving the problem as presented.

21. 21 21 Abrey

Mr Landsburg

It is time for the solution although I am not sure there is one.

If 25 is the sum of the numbers of Betty and Carol there are 12 possible combinations of numbers (1/24, 2/23, 3/22….). However, if 25 is the difference then the possible combinations are infinite.

22. 22 22 Steve Landsburg

Abrey: can you explain how 1/24 is possible?

23. 23 23 Abrey

I am sorry I don’t understand your question. Is possible as what?

However, whatever your question is, it doesn’t answer mine. Would you be kind enough to point me in the direction of the solution to the the problem as presented.

I can see how Alice, looking at 10 and 15 is able to work out how she is 25 but I can’t see how I would work out that she is looking at 10 and 15 from her Alice, Betty and Carol all declaring they don’t know what they are and then Alice declaring she is 25.

What I think I would know is that:

Alice is 25.
Neither Betty or Carol are 25
If Alice is the sum of Betty and Carol then there are 12 possible that add up to 25.
If Alice is the difference between Betty and Carol then there are infinite combinations.

24. 24 24 Steve Landsburg

Abrey: See Bennett Haselton’s comment, which contains a complete solution.

25. 25 25 deneb

So the Tripla (5,2,3) and its multiples (i.e 140,56,84) are all possible solutions for Alice to answer first. Right?