# 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 $$W$$.

• $HE×(E-1)H=W00W$ is a multiple of $11$, so $E=H+1$ Apr 14 at 16:40
• HE is a multiple of 13, so HE=78 Apr 14 at 16:44
• @Empy2 Very elegant. How do you know that HE is a multiple of 13? Apr 14 at 16:45
• From my first comment, $HE×HH=W00W$, so $HE×H=W×7×13$ Apr 14 at 16:48
• I also remember when the quickly deleted question popped up a few weeks ago. For what it's worth, I subsequently tracked the problem down to problem 7 (Division 1) of Auckland (NZ) MO 2015. May 5 at 19:20

$$HE×EH=WHEW$$. 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 $$W=6$$

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.