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

Home > Numbers
Another last digit (Posted on 2004-09-07) Difficulty: 1 of 5
What's the last digit of 11^2003 times 7^2004 times 13^2005 ?

See The Solution Submitted by Federico Kereki    
Rating: 2.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution | Comment 9 of 10 |

We know that:

7^(4a+b) = 7^b(Mod 10) and 7^(4a) = 1 (Mod 10)
3^(4a+b) = 3^b(Mod 10) and 3^(4a) = 1 (Mod 10)
1^x = 1 (Mod 10)

Thus, 11^2003 = 1 (Mod 10)
7^2004 = 1( Mod 10)
13^2005 = 3 Mod (10)

Thus,
11^2003*7^2004*13^2005
= 1*1*3(Mod 10)
= 3 (Mod 10)

Consequently, the required last digit is 3. 


  Posted by K Sengupta on 2007-07-18 01:16:36
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 (2)
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