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

Home > Just Math
Maximum Power 2 Divide Evenly (Posted on 2008-10-19) Difficulty: 3 of 5
Determine the maximum value of a positive integer G, such that (20071024 -1) is evenly divisible by 2G.

Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.

  Submitted by K Sengupta    
Rating: 4.0000 (2 votes)
Solution: (Hide)
The required maximum value of maximum value of G is 13.

EXPLANATION:

Factorizing the given expression, we have:

(2007^1024 – 1) = (2007^512 + 1) (2007^256 + 1) (2007^128 + 1) (2007^64+ 1)……..*(2007^2 + 1)*(2007+1)(2007-1) ……(i)

With the exception of (2007-1), all the other factors possess the form (20072^m +1).

For m >= 1, we observe that:

20072^m(mod 4) = (-1)2^m (mod 4) = 1
or, (20072^m+1) (mod 4) = 2

Thus, each the first 9 factors in (i) are divisible by 2, and no higher power of 2 divides these factors.

For the two remaining factors, we observe that the highest power of 2 dividing (2007-1) = 2006 is 1 and the highest power of 2 dividing (2007+1) = 2008 is 3.

Consequently, the required maximum value of maximum value of G is: (9+1+3) = 13.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionNo SubjectAdy TZIDON2008-10-19 17:31:16
Solutioncomputer solutionCharlie2008-10-19 14:36:08
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (24)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information