I believe that this is equivalent to finding an \( n \) such that the fractional parts of \( n \log 3 \) and \( n \log 7 \) are both smaller than \( \log 1.1 \). (All logs are base 10.)

It seems like this should be no trouble, since \( \log 3 \) and \( \log 7 \) are both irrational...but I haven't yet written down anything rigorous. Just thought this would help.

Yes, it easily follows from the fact that given any irrational number \(I\) and an arbitrarily small positive real \(r,\) there exist integers \(x,y\) such that
\[1>x+Iy>1-r.\]

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## Comments

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TopNewestVery very nice problem :D

Hint: Rewrite the numbers as \((5-2)^n\) and \((5+2)^n\) and use binomial theorem :D

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I don't see how the Binomial Theorem helps. Can you explain your solution?

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Look at it again. It is no coincidence that those two numbers have a mean of \(5\). :D

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Expand the first few cases

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YAY! ABSOLUTELY FANTASTIC! That's exactly what I was looking for @Patrick Corn @Mursalin Habib @Sreejato Bhattacharya! Please write more! I'm really happy that somebody's got it. :D

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I believe that this is equivalent to finding an \( n \) such that the fractional parts of \( n \log 3 \) and \( n \log 7 \) are both smaller than \( \log 1.1 \). (All logs are base 10.)

It seems like this should be no trouble, since \( \log 3 \) and \( \log 7 \) are both irrational...but I haven't yet written down anything rigorous. Just thought this would help.

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Yes, it easily follows from the fact that given any irrational number \(I\) and an arbitrarily small positive real \(r,\) there exist integers \(x,y\) such that \[1>x+Iy>1-r.\]

Or alternatively, one-liner:

:P

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Voted up for the Sonnhard reference :)

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Dr Sonnhard Graubner?

Do you knowLog in to reply

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These two links will give you a fairly good idea of how he posts. ;)

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I see what you're saying, but there's a really cool approach that doesn't use logarithms at all. :D

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Nice question/ thing to prove. I'll attempt it. BTW, do you know the proof?

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Yeah dude! This is part of a collection of proof problems I'm collecting from various olympiads. It's gonna be so boss once I've finished. :D

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Finn, im eagerly waiting for you cool approach! :)

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