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

Home > Just Math
Last Digit (Posted on 2004-01-23) Difficulty: 3 of 5
Find the last digit of summation of the series:
(1)^99 + (2)^99 + (3)^99 + (4)^99 + + (98)^99 + (99)^99

See The Solution Submitted by Ravi Raja    
Rating: 2.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): solution --- Gosh!!!.. | Comment 6 of 19 |
(In reply to re: solution by Charlie)

... I left out 3:

A number ending in 3 will have powers with ending digits that cycle 3, 9, 7, 1 in a cycle of 4. As 99 is congruent to 3 mod 4, 3^99 ends in 7, the third in the cycle.

That makes one decade add up to 5 mod 10.


  Posted by Charlie on 2004-01-23 10:22:38

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 (11)
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