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

Home > Just Math
A Remainder Problem (Posted on 2006-08-08) Difficulty: 3 of 5
Determine the positive integer values of x less than 100 such that the last two digits of 2x equal x.

What positive integer values of y less than 100 are there such that the last two digits of 3y equal y?

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution for both x and y | Comment 2 of 5 |

Hi,

x would have to be a multiple of 4, and 2^4n always ends in 6, so the only possibilities are 16, 36, 56, 76 and 96. Since the last 2 digits of powers repeat at intervals of 20, it just remains to check the last 2 digits of 2^16, which are 36.

Hence, the only possible value for x is 36.

y would have to be odd, and this would mean y ends in 3 or 7. Again inspecting the first 20 powers of 3, we find that the only possible solution for y is 87.


  Posted by nirmal on 2006-08-08 17:17:01
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information