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That is, prove that for distinct positive integers $x$, $y$, and $z$, at least one of these integers will be greater than the bitwise XOR of the other two integers.

The only "progress" I managed to make was assuming for the sake of contradiction that we could have $x<y$ XOR $z$ and so on. Then, this would imply $x+y+z<x$ XOR $y+y$ XOR $z+z$ XOR $x$. I tried proving this was impossible but later found a counterexample :(

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closed as off-topic by John Omielan, Eevee Trainer, Lee David Chung Lin, Shailesh, hardmath Mar 13 at 0:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

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    $\begingroup$ Welcome to MSE. Please give some context, in particular, tell us what you've tried so far, including anything in particular you had difficulty with. Also, letting us know where this problem comes from would be helpful. Thanks. $\endgroup$ – John Omielan Mar 12 at 20:47
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    $\begingroup$ The "bare problem statement" is disfavored, not only because it typically conveys an imperative, but also because it fails to disclose what you find interesting or difficult about the problem. Showing that you digested the problem's meaning before posting will often encourage Readers to provide the best exposition they can. $\endgroup$ – hardmath Mar 13 at 0:47
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So let's write $x$, $y$ and $z$ as binary numbers where the first digit is a $1$ (note that the numbers are strictly positive). Some hints:

  • If one number has more digits than the other two, what can we do? Can XOR ever return a number with more digits than its two inputs?
  • How can we extend this to the case where two numbers have the same number of digits as each other, and this is more than the number of digits of the third number?
  • If each number has the same number of digits, they all start with a $1$. What happens if you take the bitwise XOR of any two of the numbers?
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