BRILLIAthon $2$ (Problem $1$ )

$A^{\overline{AA}}+\overline{AA}=\overline{B,BB9,DED,BEE,BBB,BBE}$ $\implies 10^{16}>A^{\overline{AA}}+\overline{AA}>10^{15}$ Claim 1: $\;A=3$

Proof: $\;\;\;$ If $A\le 2$ , we have, $A^{\overline{AA}}+\overline{AA}\le 2^{22}+22<(2^{9})^3<(10^3)^3=10^9$ which is a contradiction. ( $A^{\overline{AA}}+\overline{AA}$ is too small)

If $A\ge 4$ , we have, $A^{\overline{AA}}+\overline{AA}\ge 4^{44}+44>(2^2)^{44}>(2^{10})^8>(10^3)^8=10^{24}$ which is again a contradiction. ( $A^{\overline{AA}}+\overline{AA}$ is too large)

Therefore, the only possibility is $\boxed{A=3}$

---

$3^{33}+33=\overline{B,BB9,DED,BEE,BBB,BBE}$ Claim 2: $\;B=5$ and $E=6$

Proof: $\;\;\;$ Since, the last two digits of $3^{33}+33$ is $\overline{BE}$ , $3^{33}\equiv 3\cdot ((3^8)^2)^2\equiv 3\cdot ((61)^2)^2\equiv 3\cdot 21^2\equiv 3\cdot 41\equiv 23 \pmod{100}$ $\implies 3^{33}+33\equiv 56 \pmod{100}$ Therefore, the only possibility is $\boxed{B=5}$ and $\boxed{E=6}$

---

$3^{33}+33=\overline{5,559,D6D,566,555,556}$ Claim 3: $\;D=0$

Proof: $\;\;\;$ Since $3$ divides $3^{33}+33$ , it must also divide the sum of the digits of $\overline{5,559,D6D,566,555,556}$ $\implies 5\cdot 9+9\cdot 1+6\cdot 4+D\cdot 2\equiv 2D \equiv 0\pmod{3}$ So, $3$ must divide $D\implies D\in \{0,3,6,9\}$ . Since $A,B,C,D$ and $E$ are all distinct digits, the only possibility is $\boxed{D=0}$ .

---

$3^{33}+33=5,559,060,566,555,556$ Therefore, $A+B+D+E=\boxed{14}$

There is no written solution, however, here's the link to a video solution:

UKMT BMO $2021$ Solutions