Find four different digits $ W, O, N, E $ that satisfies the equation

$$ \overline{NO} \times \overline{WON} \times \overline{WON} = \overline{WONEWON} $$

  • 1
    $\begingroup$ What's up with the dashes above the letters? Are we negating anything, or how should I interpret those? $\endgroup$ – Mast May 4 '19 at 19:11
  • $\begingroup$ I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number $\endgroup$ – Uvc May 4 '19 at 19:45

The answer is

W = 1, O = 3, E = 0, N = 7

$73 * 137 * 137 = 1370137$


Divide both sides by $WON$

$NO * WON = 10001 + (E000 / WON)$

$ 0 <= E <= 9$

First consider cases when $E$ is not $0$ and

$(E000 / WON) = X $

$E000 = E * 2^3 * 5^3$

$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.

Now $X$ is obviously not divisible by $5$ because

$NO * WON = 10001 + X$

So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that

$E = 0$ and

$NO * WON = 10001$

To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$


W=1, O=3, N=7, E=0

works because

73 * 137 * 137 = 1370137.



works, but I don't think that's what you meant.

  • 1
    $\begingroup$ Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :) $\endgroup$ – Rubio May 4 '19 at 5:44
  • $\begingroup$ Thanks! And oops. :/ $\endgroup$ – LarrySnyder610 May 4 '19 at 10:56

As this is not tagged "no-computers", I just used some Python :

Solution : {'W': 1, 'O': 3, 'N': 7, 'E': 0}

Method (Python) :

    def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
        for i in range(0, len(string)):
        return result
    for W in range(0, 10):
        for O in range(0, 10):
            if W == O:
            for N in range(0, 10):
                if N == W or N == O:
                for E in range(0, 10):
                    if E == W or E == O or E == N:
            # Four different digits W, O, N, E
                    # Now check whether equation is fulfilled
                    digits={"W": W, "O": O, "N": N, "E": E}
                    if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
                        print("Solution : "+str(digits))
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    $\begingroup$ I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest is isSolution w o n e = all isDigit [w,o,n,e] /\ no*won*won == wonewon /\ no != 0. $\endgroup$ – Daniel Wagner May 4 '19 at 17:29
  • $\begingroup$ You could add this as answer, too, @DanielWagner $\endgroup$ – LMD May 4 '19 at 19:56
  • $\begingroup$ (This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.) $\endgroup$ – Rubio May 4 '19 at 23:40

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