For a real number $$x$$, let $$\lfloor x \rfloor$$ be the greatest integer less than or equal to $$x$$. For example, $$\lfloor 1.7 \rfloor = 1$$ and $$\lfloor \sqrt{2} \rfloor = 1$$. Let $$N = \left\lfloor \frac{10^{93}}{10^{31}+3} \right\rfloor$$. Find the remainder when $$N$$ is divided by 100.
Solution
Put $$t = 10^{31}$$, so that $$\frac{10^{93}}{10^{31}+3} = \frac{t^3}{t+3} = t^2 - 3t + 9 - \frac{27}{t+3}$$. Since $$0 < \frac{27}{t+3} < 1$$, we get $$N = t^2 - 3t + 8$$. Both $$t^2$$ and $$3t$$ are multiples of 100, so $$N$$ leaves remainder 8 on division by 100.
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