Hello I have found a question about exclusion principle and I have love that you will help me with that question.

Prove that for each 201 number from[1,300] we can find that there is always two number that the divition of them is power of 3(except 1).

Any help will be appreciated!

  • $\begingroup$ inclusion exclusion principle,I have used it on all the number in the domaine and the final result was 189 numbers. $\endgroup$ – Gil Oct 8 '13 at 11:16
  • $\begingroup$ Can you be more specific? Also, what do you mean by "each 201 number from[1,300]"? $\endgroup$ – dfeuer Oct 8 '13 at 11:17
  • $\begingroup$ we need to prove it for any group of 201 numbers form 1 to 300 $\endgroup$ – Gil Oct 8 '13 at 11:19
  • $\begingroup$ @dfeuer: ‘Each set of $201$ integers from $[1,300]$’. $\endgroup$ – Brian M. Scott Oct 8 '13 at 11:28

HINT: Let $A$ be your set of $201$ numbers. Write each member of $A$ in the form $3^km$, where $3\nmid m$. $100$ of the integers in $[1,300]$ are multiples of $3$, so there are only $200$ possible values of $m$, but there are $201$ numbers in $A$.

  • $\begingroup$ I did it but I need the full version(it's about my final exam grade and it's really important to me that you will solve that by your self). $\endgroup$ – Gil Oct 8 '13 at 11:26
  • 4
    $\begingroup$ @Gil: I’m really not willing to go any further without some substantive contribution on your part; that hint is almost a complete solution already. $\endgroup$ – Brian M. Scott Oct 8 '13 at 11:30
  • $\begingroup$ Look I did on the exam I did a full answer , to be sure tomorrow I have to appeal , thank you for the help $\endgroup$ – Gil Oct 8 '13 at 11:38

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