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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, 8 months 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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

Expand the first few cases

- 3 years, 8 months ago

I'm afraid I still don't see it. How does expanding help?

- 3 years, 8 months 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

- 3 years, 8 months 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.

- 3 years, 8 months 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

:P

- 3 years, 8 months ago

Voted up for the Sonnhard reference :)

- 3 years, 8 months ago

I did not get that. :P

- 3 years, 8 months ago

Do you know Dr Sonnhard Graubner?

- 3 years, 8 months ago

Nope. The link says he doesn't exist. :/

- 3 years, 8 months ago

You need to log in to your AoPS account. These two links will give you a fairly good idea of how he posts. ;)

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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

- 3 years, 8 months 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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

I'll leave it to you guys for a while. :D

- 3 years, 8 months ago