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Choose any different 38 natural numbers less than 1000. Prove by using the Pigeonhole Principle that among the selected numbers there exists at least two whose difference is at most 26.

I proved an answer by increasing 26 to 27 and multiplying it by 38. Which gives an answer of 1026, but i can't use the pigeon principle to show it.

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  • $\begingroup$ Not a probability question $\endgroup$ – Henry Mar 4 '14 at 0:27
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Create $37$ pigeonholes of size $27$ - and note that $27 \times 37 =999$. Then there must be two of the $38$ numbers in the same pigeonhole. Fill in the details for a proof.

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