Given $AB\cdot F=JG$, $AD^E=AHJ$, ..., find the value of $G^E$ 



In the problem above, I know that the digits are all between 1 and 9, inclusive, at least most of them are, but I can't figure out a way other than trial and error to get the answer. Can someone please help? Thanks!
 A: To solve these types of problems, you should try to determine and check its "weakest" point at each stage. In your case, this seems to be first where you have exponentials, especially where the ratio of the number of digits of the result to that of the number being exponentiated is closer to $1$ as it seriously limits what the value of the exponential can be. In particular, consider
$$AD^{E} = AHJ \tag{1}\label{eq1A}$$
You have $E \neq 1$. Also, since $10^3 = 1000$, any $2$ digit integer of a power more than $2$ must be $4$ or more digits long. This only leaves that
$$E = 2 \tag{2}\label{eq2A}$$
Next, you can see that only $E$ and $C$ are used in one equation, so tackle that next, i.e.,
$$\begin{equation}\begin{aligned}
C^{E} & = EC \\
C^2 & = 20 + C \\
C^2 - C - 20 & = 0 \\
(C - 5)(C + 4) & = 0 \\
C & = 5
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The next one to check is the remaining exponential value, i.e.,
$$B^D = HB \tag{4}\label{eq4A}$$
Some digit $B$, which when taken to some power greater than $1$, has the same final digit. $B = 1$ is too small, and $B = 2$ is not allowed due to \eqref{eq2A}. With $B = 3$, you have $D = 5$ or $D = 9$ only, with $3^5 = 243$ already too large. With $B = 4$, you have $D \in \{3, 5, 7, 9\}$, with the only one which gives a $2$ digit result being $4^3 = 64$. $B = 5$ is not available. With $B$ being $6$ or higher, $B^D$ is at least a $3$ digit number, so the only possibility which works is
$$B = 4, D = 3, H = 6 \tag{5}\label{eq5A}$$
With most of the digits now determined, using \eqref{eq2A} and \eqref{eq5A} in \eqref{eq1A} gives
$$\begin{equation}\begin{aligned}
(10(A) + 3)^2 & = 100(A) + 60 + J \\
100(A) + 60(A) + 9 & = 100(A) + 60 + J \\
60(A) + 9 & = 60 + J
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
This gives
$$A = 1, J = 9 \tag{7}\label{eq7A}$$
You now have $7$ of the digits determined, with just $2$ left to assign in the remaining equation, i.e.,
$$AB \times F = JG \implies (14)F = 90 + G \tag{8}\label{eq8A}$$
The only multiple of $14$ between $90$ and $98$ is where $14(7) = 98$, giving
$$F = 7, G = 8  \tag{9}\label{eq9A}$$
This gives the expression value to determine to be
$$G^E = 8^2 = 64 \tag{10}\label{eq10A}$$
A: The heuristic mentioned by @John Omielan above is principally important for word problems like these (these are called cryptograms, I think), you need to limit your cases by looking for strong restrictions. The exponential is very comfortable to work with. Keeping in mind that the letters represent non-zero and distinct single digit numbers, we might proceed in a different way as:

Since $AD^E=AHJ$, we have a power of a $2$ digit number coming as a $3$ digit number, and by the very way we multiply numbers $E$ cannot be $3$ since, $AD^E$ would have to be at least of length $4$, so $\textbf{E=2}$ since it can't be $1$ either.


From the same equation possible choices for $A$ are $1,2,3$, but the squares of the $30$s begin with $9$ (i.e. more than $30^2=900$) and the squares of the $20$s begin with something $\ge 4$ by the same logic, so $\textbf{A=1}$.


Now $C^E=EC$ is the square of a digit beginning with $2$, so that $\textbf{C=5}$ is the only choice.


In $B^D=HB$, $D\ne B$ and neither can be $2$, since $2$'s already taken, so $B\ge 3$ but $3^5=243$ is of $3$ digits so $D\le 4$. So $\{B,D\}=\{3,4\}$ as unordered sets, in particular $3^4=81$, so the equation here must be $4^3=64$, giving $\textbf{B=4, D=3, H=6}$.


From what we have till now, $AD^E=AHJ$ gives $\textbf{J=9}$ and


From what we have till now, $AB\times F = JG$ gives $14\times F=90+G$ which, if you know your multiplication table of $14$ gives $\textbf{F=7, G=8}$ so that $G^E=64$.

