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[tex]\frac{18^2+2^{100}}{11}[/tex]

Now I know that if

[tex]a \equiv b\ (mod\ m),\ c \equiv d\ (mod\ m) \Rightarrow[/tex]

[tex]a + c \equiv b +d\ (mod\ m)[/tex] and [tex]ac \equiv bd\ (mod\ m)[/tex]

so

[tex]18^2 \equiv b\ (mod\ 11) \Rightarrow \frac{18^2}{11}=29.454545... \Rightarrow b=18^2-11\cdot 29=5[/tex]

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?