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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.

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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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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

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Expand the first few cases

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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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Lol i literally wrote a python program to try to figure this out I got n = 568, 1136, 2098, 2666, 2905, 4196, 4435, ...

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That's awesome!

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thanks

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Check out my new problem!

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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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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?

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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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