# 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
6 years, 3 months ago

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.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $\boxed{123}$

Sort by:

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

- 2 years, 4 months ago

That's awesome!

- 2 years, 4 months ago

Check out my new problem!

- 2 years, 4 months ago

thanks

- 2 years, 4 months ago

Very very nice problem :D

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

- 6 years, 3 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago

Expand the first few cases

- 6 years, 2 months ago

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

- 6 years, 2 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

- 6 years, 3 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.

- 6 years, 3 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

- 6 years, 3 months ago

Voted up for the Sonnhard reference :)

- 6 years, 3 months ago

I did not get that. :P

- 6 years, 3 months ago

Do you know Dr Sonnhard Graubner?

- 6 years, 3 months ago

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

- 6 years, 3 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. ;)

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago