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That seems unlikely... Prove it!

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.

Note by Finn Hulse
3 years ago

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Very 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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@Nathan Ramesh I don't see how the Binomial Theorem helps. Can you explain your solution? Jon Haussmann · 3 years ago

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@Jon Haussmann Look at it again. It is no coincidence that those two numbers have a mean of \(5\). :D Finn Hulse · 3 years ago

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@Jon Haussmann Expand the first few cases Nathan Ramesh · 3 years ago

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@Nathan Ramesh I'm afraid I still don't see it. How does expanding help? Jon Haussmann · 3 years ago

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@Nathan Ramesh 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 Finn Hulse · 3 years ago

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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. Patrick Corn · 3 years ago

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@Patrick Corn 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:

hello,

\(n=568\) works nicely.

Sonnhard

:P Sreejato Bhattacharya · 3 years ago

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@Sreejato Bhattacharya Voted up for the Sonnhard reference :) Mursalin Habib · 3 years ago

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@Mursalin Habib I did not get that. :P Finn Hulse · 3 years ago

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@Finn Hulse Do you know Dr Sonnhard Graubner? Sreejato Bhattacharya · 3 years ago

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@Sreejato Bhattacharya Nope. The link says he doesn't exist. :/ Finn Hulse · 3 years ago

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@Finn Hulse You need to log in to your AoPS account. These two links will give you a fairly good idea of how he posts. ;) Sreejato Bhattacharya · 3 years ago

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@Sreejato Bhattacharya But I am logged in. Furthermore, only one of the links you gave works. Sharky Kesa · 3 years ago

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@Patrick Corn I see what you're saying, but there's a really cool approach that doesn't use logarithms at all. :D Finn Hulse · 3 years ago

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

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@Sharky Kesa 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 Finn Hulse · 3 years ago

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@Finn Hulse Finn, im eagerly waiting for you cool approach! :) Sagnik Saha · 3 years ago

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@Sagnik Saha I'll leave it to you guys for a while. :D Finn Hulse · 3 years ago

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