Solve alphanumeric puzzle $(HE) \times (EH) = (WHEW).$ This is a self-answer query.  I came across a recent mathSE query that presented an
interesting problem, and took some time solving it.  When I  looked for the
query to respond, I couldn't find it.  I suspect that the problem may have been closed or deleted.
The real challenge is to
avoid brute force and to present a completely analytical solution.  That is why
I think it is worthwhile to present the problem and solution.
Problem
In the multiplication problem below, the letter $H$ is used to symbolize the
same digit (one of the values $0$ through $9$) throughout the problem. Similarly, the letter $E$ symbolizes a specific digit throughout the problem,
and the letter $W$ symbolizes a specific digit throughout the problem.
The letters $H,E,W$ symbolize different digits from each other.  No number is
allowed to have its leftmost digit $= 0$.  Therefore, you know immediately that none of
$H,E,$ or $W$ can equal $0$.
\begin{array}{ r  r  r  r}
 &  & H & E \\
\times &  & E & H \\
 \hline
W & H & E & W \\
\end{array}
Determine the values for $H,E,$ and $W$.
 A: $HE×EH=WHEW$.
Subtract $HE×10$ and get
$$HE×(E-1)H=W00W\\=W×7×11×13$$
H and E are different so HE is not a multiple of 11, so $(E-1)H$ is a multiple of 11, so $E=H+1$ and $(E-1)H=HH$.
$$HE×HH=W×7×11×13\\HE×H=W×7×13$$
$13$ must be a factor of $HE$, and since we know $E=H+1$, the only possible multiple of $13$ is $78$.
Lastly, $HE×H=78×7=W×7×13$, so $W=6$
A: Answer:
See also Empy2's answer, which is much more elegant than mine.

 Since $H \times E \equiv W \pmod{10},$ and $H,E,W$ are all distinct, you know that neither $H$ nor $E$ can equal $1$.  Similarly, since $W \neq 0$, you know that neither $H$ nor $E$ can equal $5$.  Therefore, you also know that $W \neq 5$.

 So, at this point, you know that $H,E \in \{2,3,4,6,7,8,9\}$ and
 that $W \in \{1,2,3,4,6,7,8,9\}.$

 You also know that $W = 9$ is impossible, because that would prevent either $H$ or $E$ from equaling $9$.  Then, the largest possible product would be $(87 \times 78) < 9000$, which would violate the constraint that the leftmost digit of the product is $W = 9$.

 Similarly, $W$ can't equal $8$, because $97 \times 79 < 8000.$

 So, at this point, you know that $W \in \{1,2,3,4,6,7\}.$

 If $W = 3$, then $H \times E \equiv W = 3\pmod{10}$.
 Therefore, $H$ and $E$ must both be $\in \{3,7,9\}.$
 The only possibility here, $H,E = 7,9$, in some order, is impossible because $(79) \times (97) > 3999.$

 So, at this point, you know that $W \in \{1,2,4,6,7\}.$

$W = 1$ is impossible, because that would require $H,E = 7,3$, in some order, and $(37) \times (73) > 1999.$

 So, at this point, you know that $W \in \{2,4,6,7\}.$

$W = 7$ is impossible, because that would require $H,E = 9,3$, in some order, and $(39) \times (93) < 7000.$

 So, at this point, you know that $W \in \{2,4,6\}.$

 Suppose that $(10 \times H) + E \equiv k \pmod{9} ~:~ k \in \{1,2,\cdots,9\}.$
 By Casting out 9's you know that $(10 \times E) + H$ and $(E + H)$ are each also $\equiv k\pmod{9}$.

 This implies that $(2W) + H + E \equiv k^2 \pmod{9} \implies $
$2W \equiv (k^2 - k) = k(k-1) \pmod{9}.$

 As $k$ ranges from $1$ through $9$, $k(k-1)$ is never equal to either $8$ or $4 \pmod{9}.$

 From this you know that $w$ can't equal either $2$ or $4$, so $w = 6$
 and $H,E \in \{2,3,4,7,8,9\}$.

 Further, since $59 \times 95 < 6000$, you know that
$H,E \in \{7,8,9\}.$

 Therefore, $H,E$ = $7,8$, in some order.
 Since $78 \times 87 = 6786$, you know that $H,E = 7,8.$ respectively.

