Tupper's inequality is \begin{equation}\frac{1}{2} \Bigl\lfloor mod\left(\Bigl\lfloor\frac{y}{17}\Bigr\rfloor 2^{-17\lfloor x \rfloor - mod(\lfloor y \rfloor, 17)},2\right)\Bigr\rfloor.\end{equation} Let's see if it holds for $x=0$ and $y=103$ Firstly, we have $$\Bigl \lfloor \frac{y}{17} \Bigr \rfloor = \Bigl \lfloor \frac{103}{17} \Bigr \rfloor = \Bigl \lfloor 6.06 \Bigr \rfloor = 6.$$ Secondly, we have $$17\lfloor x \rfloor = 17\lfloor 0 \rfloor = 0.$$ We also have $$mod(\lfloor y \rfloor, 17) = mod(\lfloor 103 \rfloor, 17) = mod(103, 17) =1.$$ Putting all this together gives $$\Bigl\lfloor mod\left(\Bigl\lfloor\frac{y}{17}\Bigr\rfloor 2^{-17\lfloor x \rfloor - mod(\lfloor y \rfloor, 17)},2\right)\Bigr\rfloor = \Bigl\lfloor mod\left(6 \times 2^{0 - 1}, 2 \right)\Bigr\rfloor =\Bigl\lfloor mod\left(\frac{6}{2}, 2 \right)\Bigr\rfloor = \Bigl\lfloor mod\left(3, 2 \right)\Bigr\rfloor = 1.$$ Since $\frac{1}{2}1$ the inequality holds for $x=0$ and $y=103$. Therefore, the square with coordinates $(0,103)$ should be coloured. Can you work out whether the square below, which has coordinates $(0,102)$, should be coloured?

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