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!
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 $A$.