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

Home > Just Math
A Power of an Integer Beginning with an Arbitrary Sequence (Posted on 2008-01-31) Difficulty: 4 of 5
L is any integer other than a power of 10. M is any integer. Show that there is an integral power of L that begins with the sequence of digits given in M.

See The Solution Submitted by FrankM    
Rating: 3.8000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Might be the solution | Comment 4 of 6 |
The Problem statement is equivalent to:
For some positive integers(x,k)
M*10^x ≤ L^k < (M+1)*10^x
Taking log for the above equation
logM + x ≤ klogL < log(M+1)+x
=> (logM+x)/logL ≤ k < (log(M+1)+x)/logL
We can find x such that
logM+x = (k-α)logL where α is approximately 0 for some k
Then log(M+1)+x = logM*(1+1/M)+x
log(M+1)+x = logM + x + ((1/M)-(1/2Mē)+...) (log(1+x) expansion)
log(M+1)+x = (k-α)logL + ((1/M)-(1/2Mē)+...)
As α is very near to 0,((1/M)-(1/2Mē)+...) will
be greater than αlogL, which makes
(log(M+1)+x)/logL is just greater than k.
Hence there exists an integral power of L which starts
with M.
 

  Posted by Praneeth on 2008-02-01 02:54:00
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (4)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information