All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Where 2^n leads, 3^n cannot be far behind... (Posted on 2012-08-24) Difficulty: 3 of 5

Carrying on in the spirit of '86 at most' here is a pair of 'statistical' conjectures about 3^n: (see A060956 in Sloane, particularly the table, for the values up to 3^1000)

To start with a definition; if the first digit of 3^n is a 9 ( e.g. 3^23 = 94,143,178,827) then we say that 3^n is 'good'.

Conjecture 1: If 3^n is good, then either 3^(n+21), or 3^(n+23) is also good.

Conjecture 2: If 3^n is good, then 3^(m+n) is also good, for some constant, m, and n greater than 2.

True or false?

See The Solution Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution -- correction Comment 2 of 2 |
(In reply to solution by Charlie)

In part 2, my analysis was a bit muddled.

The approximations should have shown closer and closer to 9, on the high side (which it did) and closer and closer to 10 but on the low side, such as 9.9999.... (some finite but lowering amount each time), which it did not.

Then the last paragraph should have said:

Any given single value for m will have its power of 3 have either a significand some finite amount above 1 or below 1 (or, rather, conventionally, below 10).  Either way, it will destroy the goodness of some number that's sufficiently close to but below 1 or close to and above 9, respectively, and this sufficient degree of closeness is always achievable.

Edited on August 24, 2012, 10:07 pm
  Posted by Charlie on 2012-08-24 22:05:37

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information