Hcfyk

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

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

The answer is

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

$73 * 137 * 137 = 1370137$

Method:

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.

Also,

W=O=N=E=0

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

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

        result=0

        for i in range(0, len(string)):

            result+=digits[string[i]]*(10**(len(string)-i-1))

        return result

    for W in range(0, 10):

        for O in range(0, 10):

            if W == O:

                continue

            for N in range(0, 10):

                if N == W or N == O:

                    continue

                for E in range(0, 10):

                    if E == W or E == O or E == N:

                        continue

            # 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))