Choosing $101$ numbers from $\{1, 2, \dots , 200\}$.

Question

I want to prove that if you choose $101$ numbers from the set $\{1,2,3,4,\dots ,200\}$, there are always two numbers such that one divides the other with no remainder. The proof should involve the "pigeonhole principle".

I am not sure how to define the pigeonholes and how to define the pigeons. Any assistance with the proof will be most appreciated.

Thank you.

Write numbers in form $2^kq$ where q is odd, then the result should follow easily — arberavdullahu, May 08 '17 at 09:05
Do you mean like this: 1=(2^0)1, 2=21, 3=(2^0)*3,...what are the holes and what are the pigeons? Not sure I am following you, — user3275222, May 08 '17 at 09:18
Well then the holes are odd number 1,3,5,..,199, then from 101 numbers there will be two with same q. — arberavdullahu, May 08 '17 at 09:23

score 11 · Accepted Answer · answered May 08 '17 at 09:45

11

Write each of the $101$ numbers as $2^kq$ for some odd $q$. There are $100$ choices for $q$. Hence there must be at least two of the $101$ numbers, $2^{k_1}p$ and $2^{k_2}p$, such that the odd number is the same and so one divides the other.

answered May 08 '17 at 09:45

Shaun

47,747

why there are 100 choices for q? I mean $101=101 \times 2^0$ – Oct 01 '19 at 16:17
Because $q$ is selected from odd divisors of numbers in ${1, 2, 3, \dots, 200}$, @LoveQYG. – Shaun Oct 01 '19 at 17:13

Choosing $101$ numbers from $\{1, 2, \dots , 200\}$.

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