Alphanumeric Puzzle multplication : $OW\times HE=WWW$

Question

$OW\times HE=WWW$ where $O,W,H,E$ are digits.

I've gathered that

$$W\in\{2,3,4,5,6,7,8,9\}$$ $$E\in\{1,3,6,7,9\}$$

but I can't seem to move past it.

$~W=9~$ (for example) can easily be rejected, by exploring the max value of the possible two digit numbers $~OW \times HE.~$ One plausible plan of attack would be to forgo elegance, and investigate possible values of $~W.~$ Note that the RHS product must be a multiple of $~3,~$ which implies that at least one of the two LHS factors must be a multiple of $~3.$ — user2661923, Apr 28 '23 at 19:38
WWW is a multiple of 111, and therefore a multiple of 37. Thus one of the 2-digit factors must be either 37 or 74 (and the other must be a multiple of 3). — Jim Ferry, Apr 28 '23 at 20:45

score 1 · Answer 1 · answered Apr 28 '23 at 21:18

$WWW$ is a multiple of $111=3×37$, so $OW$ or $HE$ has to be a multiple of $37$. Thus four cases:

$HE=37\implies7×W\equiv W\bmod10\implies 6×W\equiv0\bmod10\implies W=5\implies15×37=555$

$HE=74\implies4×W\equiv W\bmod10\implies 3×W\equiv0\bmod10\implies$ no solution, we can't have $W=0$

$OW=74\implies74×HE=444\implies$ no solution, must have $HE\ge10$ to avoid a nonzero initial digit

$OW=37\implies$ let's see if the OP can work out the second solution.

1 Answers1