Show that there exists a positive integer \(n\) such that the decimal representations of \(3^n\) and \(7^n\) both start with the digits 10.

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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 – Nathan Ramesh · 3 years ago

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– Jon Haussmann · 3 years ago

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

– Finn Hulse · 3 years ago

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

– Nathan Ramesh · 3 years ago

Expand the first few casesLog in to reply

– Jon Haussmann · 3 years ago

I'm afraid I still don't see it. How does expanding help?Log in to reply

@Patrick Corn @Mursalin Habib @Sreejato Bhattacharya! Please write more! I'm really happy that somebody's got it. :D – Finn Hulse · 3 years ago

YAY! ABSOLUTELY FANTASTIC! That's exactly what I was looking forLog 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. – Patrick Corn · 3 years ago

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Or alternatively, one-liner:

:P – Sreejato Bhattacharya · 3 years ago

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– Mursalin Habib · 3 years ago

Voted up for the Sonnhard reference :)Log in to reply

– Finn Hulse · 3 years ago

I did not get that. :PLog in to reply

Dr Sonnhard Graubner? – Sreejato Bhattacharya · 3 years ago

Do you knowLog in to reply

– Finn Hulse · 3 years ago

Nope. The link says he doesn't exist. :/Log in to reply

These two links will give you a fairly good idea of how he posts. ;) – Sreejato Bhattacharya · 3 years ago

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

– Sharky Kesa · 3 years ago

But I am logged in. Furthermore, only one of the links you gave works.Log in to reply

– Finn Hulse · 3 years ago

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

Nice question/ thing to prove. I'll attempt it. BTW, do you know the proof? – Sharky Kesa · 3 years ago

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– Finn Hulse · 3 years ago

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. :DLog in to reply

– Sagnik Saha · 3 years ago

Finn, im eagerly waiting for you cool approach! :)Log in to reply

– Finn Hulse · 3 years ago

I'll leave it to you guys for a while. :DLog in to reply