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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!

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

1 Answer 1

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 $A$.

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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). –  Gil Oct 8 '13 at 11:26
4  
@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. –  Brian M. Scott Oct 8 '13 at 11:30
    
Look I did on the exam I did a full answer , to be sure tomorrow I have to appeal , thank you for the help –  Gil Oct 8 '13 at 11:38

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