Prove that if select $101$ integers from the set $S = \{1,2,3,...,200\}$, there exist $m,n$ in the selection where $\gcd(m,n) = 1$!

Any hint how to prove this?

  • 5
    $\begingroup$ Two must be consecutive. $\endgroup$ – Robert Cardona Feb 6 '13 at 6:27
  • $\begingroup$ Divide $S$ into $100$ pairs in such a way that the members of each pair are relatively prime. $\endgroup$ – Brian M. Scott Feb 6 '13 at 6:28

Consider the partition of $\{1,2,\ldots,200\}$ into $100$ partitions $$\{1,2\},\{3,4\},\{5,6\}, \ldots,\{199,200\}$$ By pigeon-hole principle, we have two numbers in one partition say $2n-1$ and $2n$. The $\gcd$ of this pair is $1$ since if $d \vert (2n)$ and $d \vert (2n-1)$, then $d \vert (2n-(2n-1))$ i.e. $d \vert 1$. Hence, $$ \gcd(2n-1,2n) = 1$$

  • $\begingroup$ in the last step.., since d|1 , so the only possibilities of d is 1, so, gcd(2n-1,2n) = 1 , right? $\endgroup$ – chihiroasleaf Feb 6 '13 at 7:22
  • $\begingroup$ @chihiroasleaf Yes. precisely. $\endgroup$ – user17762 Feb 6 '13 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.