No won won? Wone Won

Question

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

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

Answer 1

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 $NO = 73, WON = 137$

Answer 2

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.

Answer 3

Like @ppgdev's answer, my first thought was to use algebra and focus on $E$ first. But the way I did it is sufficiently different I think it's worth giving as an alternative answer: I've edited my notation to match @ppgdev's in case anyone wants to compare the approaches.

$\overline{NO} \times \overline{WON} \times \overline{WON} - \overline{WON0000}- \overline{WON} = \overline{E000}$ factorises as $\overline{WON}(\overline{NO} \times \overline{WON} - 10000 - 1) = \overline{E000}$ so we can write $X = \overline{NO} \times \overline{WON} - 10001$, where $X = \frac{\overline{E000}}{\overline{WON}}$. Then $X + 10001 = \frac{\overline{NO} \times \overline{E000}}{X}$ implies $X^2 + 10001X = \overline{NO} \times \overline{E000}$. If we work modulo 1000 we can exploit the fact $\overline{E000}$ is a multiple of one thousand while $10001$ is one above a multiple of a thousand, and deduce $X^2 + X \equiv X(X+1) \equiv 0 \pmod{1000}$. From $X = \overline{E000} / \overline{WON}$ we see both $X \ge 0$ and $X \le 9000/100 = 90$ (we could slightly improve this upper bound by recognising the denominator is at least $102$ due to digits being distinct, but this doesn't change much). Since $0 \le X \le 90$, the only way $X(X+1)$ can be a multiple of $1000$ is if $X=0$: we can't have $X$ and $X+1$ both be multiples of five, so for $X(X+1)$ to be a positive multiple of $1000 = 2^3 \times 5^3$ would require either $X$ or $X+1$ to be divisible by $5^3 = 125$, well above our upper bound for $X$. Since $X=0$, we know $E = X/(\overline{WON} \times 1000) = 0$ and $\overline{NO} \times \overline{WON} = 10001$. The final digit of $O\times N$ must be $1$, and for $O$ and $N$ to be distinct the only possibilities are $3$ and $7$ in some order. If $\overline{NO} = 37$ then $\overline{WON} = 10001/37$ is not an integer, but if $\overline{NO}=73$ then $\overline{WON} = 10001/73 = 137$ which has the required form, with $W=1$ distinct from $E = 0$, $N = 7$ and $O = 3$ as required. Finally we check $73 \times 137 \times 137 = 1370137$.

Now that's how I actually did it, so my final steps are identical to @ppgdev. But it did involve some long divisions (or, more honestly, use of a calculator) which are a bit unsatisfactory. Can we instead establish the answer just by using digit sums?

Let's resume at the point we established $E=0$. If we work modulo nine, we can deal with the digit sum $(N + O) \times (W + O + N) \equiv 1 + 0 + 0 + 0 + 1 \equiv 2 \pmod{9}$. We know $W$, $O$, and $N$ are digits between one and nine (non-zero as they are distinct from $E$) and since $O$ and $N$ are distinct we must have $3 \le O + N \le 17$. The only multiplications that make two modulo nine are $1 \times 2$, $4 \times 5$, and $7 \times 8$ because most of $2$, $11$, $20$, $29$, $38$, $47$, $56$, $65$, $74$ are either prime or have a two-digit prime factor. The possibilities for $N+O$ and $W+O+N$ are tabulated below, and show either $W=1$ with $N+O$ one of $\{4, 7, 10, 13, 16\}$, or $W=8$ with $N+O$ in $\{5, 8, 11, 14, 17\}$.

N + O (mod 9)	N + O	W + O + N (mod 9)	(W+O+N) - (N+O) (mod 9)	W
1	10	2	1	1
2	11	1	-1	8
4	4 or 13	5	1	1
5	5 or 14	4	-1	8
7	7 or 16	8	1	1
8	8 or 17	7	-1	8

If we work modulo 11, we get the alternating digit sum instead. Take units as positive, tens negative, etc: $(-N + O) \times (W - O + N) \equiv 1 - 0 + 0 - 0 + 1 \equiv 2 \pmod{11}$. Since $O$ and $N$ are distinct digits between one and nine, $O-N$ must be a non-zero integer between $-8$ and $8$ inclusive. Let's write $Y \equiv O - N \pmod{11}$. Trying $W=8$ we get $Y(8-Y)\equiv 2 \pmod{11}$, so $Y^2 - 8Y + 2 \equiv 0 \pmod{11}$. A good way to solve quadratic equations in modular arithmetic is to complete the square: $(Y-4)^2 \equiv 16 - 2 \equiv 3 \pmod{11}$. Now, essentially following the steps of the quadratic formula, we find the square roots of three mod eleven. There are algorithms for this (Tonelli-Shanks, Cipolla's, Pocklington's) but mod eleven it's easy to just check all possibilities: the only ones that work are $Y-4 \equiv 5, 6 \pmod{11}$ so $Y \equiv 9, 10 \pmod{11}$. The only values of $O-N$ in our required range are $-2$ and $-1$. This means $N>O$, so $N$ is at least two. This can't work, as then $\overline{NO} \times \overline{WON}$ is over $20 \times 800 = 16000$, far above the required $10001$. (In fact we could have avoided solving a quadratic equation modulo eleven if we'd noticed eight looked suspiciously high for $W$. Even $13 \times 800 = 10400$ is too large, so we needed $\overline{NO} < 13$. For distinct digits, the only possibility that remains is $\overline{NO} = 12$ but clearly $12 \times 821 \neq 10001$.) We must have $W=1$ and $Y(1-Y)\equiv 2 \pmod{11}$. A nice trick is that $Y^2 - Y$ is the same as $Y^2 - 12Y$ (since we're allowed to add or subtract any multiple of eleven) which is easier to complete the square on: $Y^2 - 12Y + 2 \equiv 0 \pmod{11}$ gives $(Y-6)^2 \equiv 36 - 2 \equiv 1 \pmod{11}$. The only possibilities are $Y-6 \equiv -1,1 \pmod{11}$ so $Y \equiv 5,7 \pmod{11}$. Don't like the idea of completing the square to solve a quadratic in modular arithmetic? Rewrite the equation as $Y(Y-1) \equiv -2 \equiv 9 \equiv 20 \equiv 31 \equiv 42 \equiv \cdots \pmod{11}$. Then by observation, $Y(Y-1)$ is $5 \times 4 = 20$ or $7 \times 6 = 42$. This still needs me to use a fact about quadratic equations in modular arithmetic: with an odd, prime modulus (so eleven is fine, but working modulo nine is more complicated), if we've found two solutions then we've found them all.

N\O	2	3	4	5	6	7	8	9
2	S.	..	..	S.	..	.D	S.	.D
3	..	..	S.	..	..	S.	.D	..
4	..	S.	..	..	S.	..	..	SD
5	S.	..	..	S.	..	..	S.	..
6	.D	..	S.	..	..	S.	..	..
7	..	SD	..	..	S.	..	..	S.
8	SD	..	.D	S.	..	..	S.	..
9	..	.D	S.	.D	..	S.	..	..

Fine, I hear you cry, digit sums let you avoid any long division. They're clearly a powerful tool. But you still had three long multiplications to check at the end — and why didn't you consider the final digit? Show me the lazy method! I don't want no stinking tables! And definitely, definitely, no quadratics!

So here's the "easy" method from combining the best of these approaches. Let's skip to the point where we know $E=0$ and $\overline{NO} \times \overline{WON} = 10001$. Working modulo ten (final digit check), $O \times N \equiv 1 \pmod{10}$. The only possible multiplications that achieve this are $2 \times 2$, $3 \times 7$ and $9 \times 9$ (since most of $1$, $11$, $21$, $31$, $41$, $51$, $61$, $71$, $81$ are prime or have a two-digit prime factor) but $O$ and $N$ must be distinct, so they must be $3$ and $7$ in some order. Whichever order, we know $O+N=10$ and $Y = O-N = \pm 4$. Working modulo nine (digit sum), $(N + O)(W + O + N) \equiv 2 \pmod{9}$ simplifies to $10(W + 10) \equiv 1(W+1) \equiv 2 \pmod{9}$ so we know $W \equiv 1 \pmod{9}$. In the valid range of digits, this can only be $W = 1$. Working modulo eleven (alternating digit sum), $(-N + O)(W - O + N) \equiv 2 \pmod{11}$ simplifies to $Y(1-Y) \equiv 2 \pmod{11}$. Trying $Y=-4$ gives $-4(5) \equiv -20 \equiv 2 \pmod{11}$ which works, but $Y=4$ gives $4(-3) \equiv -12 \equiv 10 \pmod{11}$ which doesn't. So we must have $O+N=10$ and $Y=O-N=-4$, hence $O=(10+(-4))/2=3$ and $N=(10-(-4))/2=7$. The only possibility we need to test is therefore $73 \times 137 = 10001$ and the only solution is $73 \times 137 \times 137 = 1370137$.

Answer 4

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))