Let integer $N=\left\lfloor{(\sqrt{29}+\sqrt{21})^{2020}}\right\rfloor$ where $\lfloor{x}\rfloor$ denotes the largest integer not exceeding $x$. Find the last two digits of $N$.
Let $A=\sqrt{29}+\sqrt{21}$ and $B=\sqrt{29}-\sqrt{21}$. Then, we have $0 < B < 1$. Meanwhile, it can be shown by binomial expansion that the value of $\left(A^{2020} + B^{2020}\right)$ is an integer (see below). Therefore, it must be true that $N= A^{2020} + B^{2020} - 1$. $$\begin{array}{ll} & A^{2020}+B^{2020} =(\sqrt{29}+\sqrt{21})^{2020} + (\sqrt{29}-\sqrt{21})^{2020} \\ =& \displaystyle 2\times\left(\binom{2020}{0}\left(\sqrt{29}\right)^{2020}+\binom{2020}{2}\left(\sqrt{29}\right)^{2018}\left(\sqrt{21}\right)^{2}+\cdots \right. \\ &\left. +\displaystyle\binom{2020}{2020}\left(\sqrt{21}\right)^{2020}\right) \end{array}$$
Because every term on the right is an integer, so their sum is an integer too.
Meanwhile, we have $$\begin{array}{ll} &A^{2020}+B^{2020} = (50 + 2\sqrt{29}\sqrt{21})^{1010} + (50 - 2\sqrt{29}\sqrt{21})^{1010} \\ =& \displaystyle 2\times\left(\binom{1010}{0}\cdot 50^{1010}+ \cdots + \binom{1010}{2}\cdot 50^2\cdot\left(2\sqrt{29}\sqrt{21}\right)^{1008}\right. \\ & \left. +\displaystyle \binom{1010}{1010}\left(2\cdot\sqrt{29}\sqrt{21}\right)^{1010}\right) \end{array}$$
Every term except the last one is a multiple of $100$. Therefore, $$\begin{array}{ll} N&= A^{2020} + B^{2020} -1 \\ &\equiv 2\cdot \binom{1010}{1010}\left(2\cdot\sqrt{29}\sqrt{21}\right)^{1010} - 1 \\ &\equiv 2^{1011}\cdot 3^{505}\cdot 7^{505}\cdot 29^{505} - 1\\ &= \boxed{51} \pmod{100}\end{array}$$
The last step utilizes the "a quick way to find the tens digit" techniques as described in the book Practice by Subject - Number Theory (MOD).