Find the remainder of the equation

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To find the remainder of the equation (18^2 + 2^100) mod 11, the first step is calculating 18^2 mod 11, which results in a remainder of 5. The challenge arises with 2^100, which is too large for direct computation. Utilizing properties of modular arithmetic, specifically that a^k mod m can simplify calculations, is suggested. By applying Fermat's Little Theorem, 2^10 mod 11 can be computed, leading to a more manageable calculation for 2^100 mod 11. The final remainder can then be determined by combining the results of both components.
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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?
 
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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
 
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