Sum of last two digit

${ 2 }1^{ 5 }\equiv 1\quad \left( \text{mod 100} \right)$

${ 21 }^{ 20001 }\equiv { \left( { 21 }^{ 5 } \right) }^{ 4000 }\cdot 21\equiv 21\quad \left( \text{mod 100} \right)$

Sum of the last 2 digits is $2+1 = 3$ .

$\displaystyle 21^{20001}$

$\displaystyle = (20+1)^{20001}$

$\displaystyle = \displaystyle \sum_{r=0}^{20001} {20001\choose r} \cdot 20^{20001-r} \cdot 1^{r}$

$\displaystyle = {20001 \choose 0} \cdot 20^{20001-0} \cdot 1^0 + \ldots \ldots + {20001\choose 20000} \cdot 20^{20001-20000}\cdot 1^{20000} + {20001\choose 20001} \cdot 20^{20001-20001} \cdot 1^{20001}$

$\displaystyle = \ldots + (20001)\cdot 20^{1} \cdot 1^{20000} + (1)\cdot 20^{0}\cdot 1^{20001}$

$\displaystyle = \ldots + 400020 + 1$

$\displaystyle = \ldots + \ldots + 21$

We know that the terms before these are multiples of 100 and won't contribute to the last 2 digits of the number.

Hence, the last two digits are 2 and 1 and their sum is $\displaystyle \boxed{3}$ .

Finding the last two digits is equivalent to getting $\displaystyle 21^{20001} \text{ (mod 100)}.$ Since $100 = 25 \times 4$ and $\displaystyle \text{gcd(25, 4) = 1}$ , using the Chinese Remainder Theorem , we have $\displaystyle 21^{20001} \text{ (mod 25)}$ and $\displaystyle 21^{20001} \text{ (mod 4)} \equiv \color{#3D99F6}{1} \text{ (mod 4)}.$ Now, working on the first one, we first note that $\displaystyle \phi(25) = 20$ and so the exponent of $21$ is reduced to $\displaystyle 20001 \text{ (mod 20)} \equiv 1 \text{ (mod 20)}$ . Thus, that is just simply $\displaystyle 21^1 \text{ (mod 25)} \equiv \color{#D61F06}{21} \text{ (mod 25)}.$ We are left with solving the equation $\displaystyle 25a + \color{#D61F06}{21} = 4b + \color{#3D99F6}{1}$ $\displaystyle \forall$ $\displaystyle a, b \ge 0,$ $\displaystyle \in \mathbb Z.$ Starting from $a = 0$ , we see that $b = 5$ with both sides equivalent to $21$ . Hence, the last two digits are $2, 1$ and their sum is $3.$