Question

(Assignment 4 - Strong Induction, Pigeon Hole Principle, Combinations and Permutations)

Prove that if n + 1 integers are selected from {1, 2, …, 2n}, then the selection
includes integers a and b such that a divides b (that is there exists an integer k such that
ak = b).

Answer 1

Let the pigeons be the selected n+1 integers.

n pigeonholes are defined which corresponds to the odd integers 1, 3, 5,.. 2n-1.

every selected integer is placed into the pigeonhole based on its largest odd divisor (which has to be one among 1, 3, 5,.. 2n-1).

if integer 'x' is placed inside the pigeonhole m ( m is the largest odd divisor of x)

=> x = 2^km for some integer k ≥ 0.

as there are n integers placed inside n-1 pigeonholes,

= some pigeonhole contains two integers a and b, where a < b.

if this pigeonhole represents odd integer t,then, a = 2^rt and b = 2^st, where r < s, so that a*2^s-t = b.

= s-r is a positive integer => follows a | b.