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

Home > Numbers
Number power (Posted on 2006-12-30) Difficulty: 3 of 5
Determine the last two and the first two digits of:

32^1+32^2+32^3+...+32^2007

No Solution Yet Submitted by atheron    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
my two cents Comment 4 of 4 |
Let x be the number in question. x=(32^2008 - 32)/31. Since the first two digits are unaffected by the -32 term, consider x to be approximately 32^2008/31. Taking logs and simplifying yields log x = 3020.849795 --> x = 7.076 X 10^3020 so the first two digits are 70. Also 31x = (32^4)^502 - 32 --> 31x=1048576^502 - 32 --> 31x=76^502 - 32 mod100 --> 31x=76-32 mod100 (since 76^k = 76 mod100). So 31x=44 mod100 --> x=24 mod100 --> last two digits are 24.
  Posted by Dennis on 2006-12-30 17:37:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information