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

Home > Numbers
Divisible by 2007 (Posted on 2007-03-11) Difficulty: 3 of 5
Prove that A= 3410^n +3000^n -2964^n -1439^n is a multiple of 2007 for all positive integer values of n.

See The Solution Submitted by Dennis    
Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution : With A Different Methodology Comment 3 of 3 |

We observe that:
A = 3410^n +3000^n -2964^n -1439^n
= (3410^n -2964^n)  - (3000^n-1439^n)
Now, (3410^n -2964^n) is divisible by 3410 -2964 = 446;
while, (3000^n-1439^n) is divisible by 3000-1439 =1561
Accordingly, A is divisible by GCD(446, 1561) = 223

Again, A = 3410^n +3000^n -2964^n -1439^n
= (3410^n -1439^n) + (3000^n-2964^n)
Now, (3410^n -1439^n) is divisible by 3410 -1439 = 2961;
while, (3000^n - 2964^n) is divisible by 3000-2469 =531
Accordingly, A is divisible by GCD(531, 2961) = 9

Consequently, A is divisible by LCM(223,9) = 223*9 = 2007, whenever A is a positive integer.


  Posted by K Sengupta on 2007-03-14 03:41:32
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