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The statements of the following two puzzles are almost identical. The only difference consists in the words printed in boldface. The two solutions, however, turn out to be totally different.

Variant 1 of the puzzle

My friends Xavier, Yvo, and Zeno are excellent mathematicians and always think strictly logically. Yesterday I told them: "I have secretly chosen three positive integers $x,y,z$ with $x+y+z=2015$. I have told $x$ to Xavier, $y$ to Yvo, and $z$ to Zeno."
Then the following conversation developed.

  1. Xavier said: I know that Yvo and Zeno have different numbers.
  2. Yvo said: I already knew that all our numbers are different.
  3. Zeno said: Aha! Now I know all three numbers.

Determine the values of $x$, $y$, and $z$!

Variant 2 of the puzzle

My friends Xavier, Yvo, and Zeno are excellent mathematicians and always think strictly logically. Yesterday I told them: "I have secretly chosen three positive integers $x,y,z$ with $x+y+z=2015$. I have told $x$ to Xavier, $y$ to Yvo, and $z$ to Zeno."
Then the following conversation developed.

  1. Xavier said: I know that Yvo and Zeno have different numbers.
  2. Yvo said: Aha! Now I know that all our numbers are different.
  3. Zeno said: Aha! Now I know all three numbers.

Determine the values of $x$, $y$, and $z$!

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  • $\begingroup$ Should we guess the numbers x, y, and z or should we determine how Zeno guessed them? $\endgroup$
    – dmg
    Jan 28, 2015 at 16:53

1 Answer 1

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X's statement means that his number must be even (if it was odd, Y and Z would have numbers summing to an even number, so they could be the same).

In the first case, Y's statement means that Y has an even number, and neither X nor Z can have the same even number. Therefore Y must have a number that is at least 1008. Z knows all three numbers, so X and Y must both have the smallest possible numbers, 2 and 1008 respectively. (Any other sum for X and Y would allow multiple possibilities for their numbers.) This leaves Z with 1005.

In the second case, Y did not know that the numbers were different before, but did after hearing X's statement. Y must not have an even number greater than 1006 (because he would have known already). Y cannot have an even number less than 1008 either, because there would be no way for him to tell that it wasn't the same as X's number. Y also can't have an odd number which is 3 mod 4, because then X and Z could have the same even number. Therefore Y must have a number which is 1 mod 4.

Now Y must have 1, otherwise it would be possible to subtract 4 from Y and add 4 to X without Z being able to tell. X must have 2 or 4. Z has 2012 or 2010. But if Z had 2012, he would have known all the numbers after X's statement. If his statement means that he did not know until Y's statement, the numbers must be 4, 1, and 2010.

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  • $\begingroup$ +1, i was just working this out myself. I had only gotten as far as figuring out odd/even and that Y would be >/< half. $\endgroup$
    – Scimonster
    Jan 28, 2015 at 17:11
  • $\begingroup$ How long did it take you to solve that? Had you seen similar ones before? $\endgroup$
    – smci
    Sep 22, 2015 at 20:54
  • $\begingroup$ i wonder if there is an algebraic proof for both the variants. $\endgroup$
    – Craftsman
    Jun 25, 2020 at 18:06

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