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No, X and Y have to be 20 and 30. Here's why:
Random Scenario 1
Say X is 10 and Y is 40 (Z is 50).
First, Z sees 10 and 40. He knows he can be 50 or 30, so he is unsure.
Next, X sees 40 and 50. He knows he can be 90 or 10, so he is unsure.
Next, Y sees 10 and 50. He knows he can be 40 or 60, so he is unsure.
Now Z assumes he was 30. In that case, X would see 40 and 30 and would be unsure of whether he was 70 or 10 (which checks out since he was unsure). Y would see 10 and 30 and would be unsure of whether he was 20 or 40 (which also checks out).
Now Z assumes he was 50. In that case, X would see 40 and 50 and would be unsure (10/90), and Y would see 10 and 50 and would also be unsure (40/60), which also checks out.
So Z doesn't have a way to know whether he's 30 or 50.
Proper Scenario
X is 20, Y is 30, Z is 50
First, Z sees 20 and 30. He doesn't know whether he's 50 or 10.
Next, X sees 30 and 50. He doesn't know whether he's 80 or 20.
Next, Y sees 20 and 50. He doesn't know whether he's 30 or 70.
Now Z assumes he's 10. In that case, X would see 10 and 30. He would be unsure of whether he was 20 or 40. Y would see 20 and 10. He could be 10 or 30, but if he was 10, he would violate the rules since both Z and Y would be 10. The only number he could be is 30, and he would have been sure of it first time through.
From this, Z knows he's 50 since if he was 10, Y would be sure of his number (30) on the first pass.
As far as I can tell, 20, 30 and 50 are the only numbers for which this works. Multiples would work too (50, 75, 125), but then the order would be screwed up.
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