# Find the remainder of the equation

1. Jan 24, 2007

### parsifal

The task is to find the remainder of the equation:
$$\frac{18^2+2^{100}}{11}$$

Now I know that if
$$a \equiv b\ (mod\ m),\ c \equiv d\ (mod\ m) \Rightarrow$$
$$a + c \equiv b +d\ (mod\ m)$$ and $$ac \equiv bd\ (mod\ m)$$

so

$$18^2 \equiv b\ (mod\ 11) \Rightarrow \frac{18^2}{11}=29.454545... \Rightarrow b=18^2-11\cdot 29=5$$
and d<6 as the remainder b+d < 11.

But as 2^100 is so large, I can't find d the way I found b. How to find it, or is there some other more convenient way that doesn't involve separating 18^2 and 2^100?

2. Jan 24, 2007

### mjsd

have you tried using this $$a\equiv b (\text{mod}\; m) \Rightarrow a^k\equiv b^k (\text{mod}\; m)$$ to help?

The answer should be obvious after the use of this theorem