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Exercise 1100Combinatorics › Pigeonhole principleOlympiadAdded 3 Nov 2026

Many odd numbers

#pigeonhole#odd#sums

Statement

Show that among any n+1n+1 integers chosen from {1,…,2n}\{1,\ldots,2n\}, at least ⌈(n+1)/2⌉\lceil (n+1)/2\rceil have the same parity.

Hint

22 parities, n+1n+1 integers.

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Solution

Pigeonhole. n+1n+1 integers split into 22 parity classes: one has ≥⌈(n+1)/2⌉\ge\lceil(n+1)/2\rceil.

Conclusion. At least ⌈(n+1)/2⌉\lceil (n+1)/2\rceil have the same parity.