A tricky problem by Bs fire

Let $a=10^{31}$ and

$\begin{aligned} N & = \left \lfloor \frac {10^{93}}{10^{31}+3} \right \rfloor = \left \lfloor \frac {a^3}{a+3} \right \rfloor \\ & = \left \lfloor \frac {(a+3)^3 - 9a^2 - 27a - 27}{a+3} \right \rfloor \\ & = \left \lfloor \frac {(a+3)^3 - 9(a+3)^2 + 27(a+3) - 27}{a+3} \right \rfloor \\ & = \left \lfloor (a+3)^2 - 9(a+3) +27 - \frac {27}{a+3} \right \rfloor \\ & = (a+3)^2 - 9(a+3) +26 \\ & = a^2 - 3a + 8 \end{aligned}$

Then $N \equiv a^2 - 3a + 8 \equiv \boxed 8 \text{ (mod 100)}$ .

• To start with, I’m going to note that one can expand $\dfrac{a}{b+c}$ to $\dfrac{a}{b}-c\dfrac{\frac{a}{b}}{b+c}$ . This may not seem all that useful but it’s super helpful for approximating stuff.
• Next, if we want the last two digits of the number, we’re going to have to consider $\pmod {100}$ and thus we can nix any multiples of $100$ .
• Lastly, GIF or [▪] is the floor function, so we’ll be rounding down at the end.

So, with that in mind:

$\begin{aligned} \left[\dfrac{10^{93}}{10^{31}+3}\right] & = \left[\dfrac{10^{93}}{10^{31}} - 3\dfrac{\frac{10^{93}}{10^{31}}}{10^{31}+3}\right] \\ & = \left[10^{62} - 3\dfrac{10^{62}}{10^{31}+3}\right] \\ & = \left[- 3\dfrac{10^{62}}{10^{31}+3}\right] \pmod{100} \end{aligned}$

This is basically in the same form as the original fraction, so we can do it again:

$\begin{aligned} \left[-3 \dfrac{10^{62}}{10^{31}+3}\right] & = \left[-3 \dfrac{10^{62}}{10^{31}} + 9\dfrac{\frac{10^{62}}{10^{31}}}{10^{31}+3}\right] \\ & = \left[-3 \times 10^{31} + 9\dfrac{10^{31}}{10^{31}+3}\right] \\ & = \left[9\dfrac{10^{31}}{10^{31}+3}\right] \pmod{100} \end{aligned}$

Here we note that $\dfrac{10^{31}}{10^{31}+3}$ is ever so slightly less than $1$ , from which $9\dfrac{10^{31}}{10^{31}+3}$ is ever so slightly less than $9$ and thus rounds down to $\fbox{8}$ .