A combinatorics problem from pre RMO 2015

A subset $B$ of the set of first 100 positive integer has the property that no two elements of $B$ sum upto 125. What is the maximum number of elements in $B$ ?

#Combinatorics

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Maximum number of elements in R is 62. From 1 to 24 and either one of (62, 63), (61, 64), (60, 65), . . . . ., (26, 99), (25, 100).

Rajen Kapur - 5 years, 6 months ago

Yes , correct. The basic idea is complementation. We find $a,b$ such that $a+b=125$ and we have 38 pairs of it. So excluding such pairs , the remaining numbers that 1 to 24 must at least be included in set $B$ . To make number of elements maximum , we include either all $a$ or $b$ making the max. number of elements as $24+38=62$ .

Nihar Mahajan - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$