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.

Posted by ken on 2007-03-11 14:07:25