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
\[1>x+Iy>1-r.\]

## Comments

Sort by:

TopNewestVery very nice problem :D

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

Log in to reply

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

Log in to reply

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

Log in to reply

Expand the first few cases

Log in to reply

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

Voted up for the Sonnhard reference :)

Log in to reply

Log in to reply

Dr Sonnhard Graubner?

Do you knowLog in to reply

Log in to reply

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

You need to log in to your AoPS account.Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply