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.

#NumberTheory #Finn #Digits #CanYouProveIt?

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Very very nice problem :D

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

Nathan Ramesh - 7 years, 1 month ago

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 - 7 years, 1 month ago

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

Jon Haussmann - 7 years, 1 month ago

Expand the first few cases

Nathan Ramesh - 7 years, 1 month ago

@Nathan Ramesh – I'm afraid I still don't see it. How does expanding help?

Jon Haussmann - 7 years, 1 month ago

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

Finn Hulse - 7 years, 1 month ago

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

Stefan Popescu - 3 years, 2 months ago

That's awesome!

Finn Hulse - 3 years, 2 months ago

thanks

Stefan Popescu - 3 years, 2 months ago

Check out my new problem!

Stefan Popescu - 3 years, 2 months ago

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

Sharky Kesa - 7 years, 1 month 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. :D

Finn Hulse - 7 years, 1 month ago

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

Sagnik Saha - 7 years, 1 month ago

@Sagnik Saha – I'll leave it to you guys for a while. :D

Finn Hulse - 7 years, 1 month ago

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 - 7 years, 1 month ago

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

Sreejato Bhattacharya - 7 years, 1 month ago

Voted up for the Sonnhard reference :)

Mursalin Habib - 7 years, 1 month ago

@Mursalin Habib – I did not get that. :P

Finn Hulse - 7 years, 1 month ago

@Finn Hulse – Do you know Dr Sonnhard Graubner?

Sreejato Bhattacharya - 7 years, 1 month ago

@Sreejato Bhattacharya – Nope. The link says he doesn't exist. :/

Finn Hulse - 7 years, 1 month ago

@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 - 7 years, 1 month ago

@Sreejato Bhattacharya – But I am logged in. Furthermore, only one of the links you gave works.

Sharky Kesa - 7 years, 1 month ago

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

Finn Hulse - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$