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Exercise 525Combinatorics › Pigeonhole principleHigh schoolAdded 3 Nov 2026

Among $n+1$ from $2n$, two consecutive

#pigeonhole#consecutive#betweenness

Statement

Show that among any n+1n+1 integers from {1,…,2n}\{1,\ldots,2n\}, some two are consecutive.

Hint

Group into nn pairs (1,2),(3,4),…(1,2),(3,4),\ldots.

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Solution

Pairs. nn pairs (2k−1,2k)(2k-1,2k) for k=1,…,nk=1,\ldots,n.

Pigeonhole. n+1n+1 integers in nn pairs: some pair has 22.

Conclusion. Two are consecutive.