2014/085) prove that $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 2000

proof:

We have 2000 = 125 * 16
So it should be divisible by 125 * 16 or 125 and 16 as these are co-primes

$a^n-b^n$  is divisible by $a-b$
hence
$121^n – 25^n$ is divisible by $121- 25 = 96= 16 * 6$ hence 16
$1900^n – (-4)^n$ is divisible by $1900- (-4) = 1904 = 16 * 1 19$ hence 16
So $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 16

$121^n – (-4)^n$ is divisible by $121 + 4 = 125$
$– 25^n + 1900^n$ is divisible by $1900 – 25 = 1875 = 125 * 15$ hence 125
So $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 125

As the number is divisible by 125 and 16 hence the product that is 2000