Prove that A= 3410^n +3000^n -2964^n -1439^n is a multiple of 2007 for all positive integer values of n.
(In reply to
Spoiler Ahead ! by K Sengupta)
Just a public thank you to K Sengupta for still leaving enough work left with his/her spoiler to make the problem both fun and educational for me!
A= 3410^n +3000^n -2964^n -1439^n
3410 mod 223 = 65
3000 mod 223 = 101
2964 mod 223 = 65
1439 mod 223 = 101
So,
A mod 223 = 65^n + 101^n - 65^n - 101^n mod 223 = 0 mod 223
also,
3410 mod 9 = 8
3000 mod 9 = 3
2964 mod 9 = 3
1439 mod 9 = 8
A mod 9 = 8^n + 3^n - 3^n - 8^n mod 9 = 0 mod 9
meaning A is integrally divisible by both 223 and 9 (and, hence, 2007) regardless of n.
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Posted by ken
on 2007-03-11 14:07:25 |
re: Spoiler Ahead ! (public thank you with full solution)
