JBMO Shortlist

Solution:

Denote
$$x=\overline{V E R}-\overline{I A}, \quad y=\overline{G R E}+\overline{E C E}, \quad z=G^{R^{E}}$$
Then obviously, we have
$$\begin{aligned} & (201+131 \text{ or } 231+101) \leq y \leq (879+969 \text{ or } 869+979 \text{ or. } 769+989) \\ \Rightarrow \quad & 332 \leq y \leq 1848 \Rightarrow 102-98 \leq x \leq 987-10 \Rightarrow 4 \leq x \leq 977, \end{aligned}$$
hence it follows that
$$\frac{4}{1848} \leq \frac{x}{y}=z \leq \frac{977}{332} \Rightarrow 1 \leq z \leq 2$$
This shows that $z=G^{R^{E}} \in\{1,2\}$. Hence, if $R \geq 1$, then $R^{E} \geq 1$, which implies that $2 \geq G^{R^{E}} \geq G$. Thus, if $R \geq 1$, then it must be $G \leq 2$. In view of this and the assumption of the problem that the number $\overline{G R E E C E}$ has a maximum value, we will consider the case when $R=0$ hoping to get a solution with $G>2$. Then $G^{R^{E}}=G^{0}=1$ for all digits $G$ and $E$ with $1 \leq G, E \leq 9$, and therefore, the above equality becomes
$$\overline{V E R}-\overline{I A}=\overline{G R E}+\overline{E C E}$$
which substituting $R=0$, can be written as
$$\overline{V E O}=\overline{G O E}+\overline{E C E}+\overline{I A}$$
Now we consider the following cases:

Case 1: $G=9$. Then $V \leq 8$, so $\overline{V E 0} \leq 900$, while the right hand side of (1) is greater than 900. This is impossible, and no solution exists in this case.

Case 2: $G=8$. Then (14) becomes
$$\overline{V E 0}=\overline{80 E}+\overline{E C E}+\overline{I A}$$
hence it immediately follows that $V=9$. For $V=9$, (15) becomes
$$\overline{9 E 0}=\overline{80 E}+\overline{E C E}+\overline{I A}$$
Notice that for $E \geq 2$, the right hand side of (16) is greater than 1000, while the left hand side of (16) is less than 1000. Therefore, it must be $E \leq 1$, that is, $E=1$ in view of the fact that $R=0$. Substituting $E=1$ into (16), we get
$$\overline{910}=\overline{801}+\overline{1 C 1}+\overline{I A}$$
hence it follows that
$$109=\overline{1 C 1}+\overline{I A}$$
But the right hand side of (18) is greater than 121. This shows that $G=8$ does not lead to any solution.

Case 3: $G=7$. Then (14) becomes
$$\overline{V E O}=\overline{70 E}+\overline{E C E}+\overline{I A}$$
Thus it must be $V \geq 8$.

Subcase 3(a): $V=8$. Then (19) gives
$$\overline{8 E 0}=\overline{70 E}+\overline{E C E}+\overline{I A}$$
hence we immediately obtain $E=1$ (since the right hand side of (6) must be less than 900). For $E=1$, (20) reduces to
$$109=\overline{1 C 1}+\overline{I A}$$
which is impossible since $\overline{1 C 1} \geq 121$.

Subcase 3(b): $V=9$. Then (19) gives
$$\overline{9 E 0}=\overline{70 E}+\overline{E C E}+\overline{I A}$$
hence we immediately obtain $E \leq 2$ (since the right hand side of (7) must be less than 1000). For $E=2$, (22) reduces to
$$218=\overline{2 C 2}+\overline{I A}$$
which is impossible since $\overline{2 C 2} \geq 232$. Finally, for $E=1$, (22) reduces to
$$209=\overline{1 C 1}+\overline{I A}$$
Since it is required that the number $\overline{G R E E C E}$ has a maximum value, taking $C=8$ into (24) we find that
$$28=\overline{I A}$$
which yields $8=A=C$. This is impossible since must be $A \neq C$. Since $C \neq G=7$, then taking $C=6$ into (24) we obtain
$$48=\overline{I A}$$
hence we have $I=4$ and $A=8$. Previously, we have obtained $G=7, R=0, V=9, E=1$ and $C=6$. For these values, we obtain that $\overline{G R E E C E}=701161$ is the desired maximum value.