What is the remainder when $$128^{1000}$$ is devided by 153?

$$128^{1000}$$ can be written as $$2^{7000}$$ and the divisor i.e. 153 can be broken down into 9*17. 
$$2^{7000}$$ can be written as $$8^{2333}\times\ 2$$. 8 leaves -1 as remainder when divided by 9. So, $$8^{2333}\times\ 2$$ will leave $$-1^{2333}\times\ 2$$ as remainder i.e. -2 when divided by 9. So, the remainder when the number is divided by 9 is 9-2 = 7.

Similarly, $$2^{7000}$$ can be written as $$16^{1750}$$ and 16 leaves -1 remainder when divided by 17 so $$16^{1750}$$ will also leave $$-1^{1750\ }$$ i.e. 1 remainder when divided by 17.

Now, the remainder can be written as: N = 9a + 7 = 17b + 1.
17b - 9a = 6. Possible values for a and b are 5 and 3 rsepctively.
Thus, N = 52 and the number can be written as 153K + 52 thus leaving 52 as remainder when divided by 153.