# French alphametic (corrected)

This is the corrected version of puzzle French alphametic (misspelled)

Every letter stands for a digit in base-10 representation, different letters stand for different digits, and the four summands and the sum are even:

         UN
UN
DEUX
+  DOUZE
------------
SEIZE


Which digit does each letter represent?
(Please present the full analysis how these digits can be determined.)

• good now there are 12 solutions :) – Oray Feb 11 '16 at 13:25
• @Oray Now there are 2 solutions - he added an additional restriction that all the numbers are even. – Trenin Feb 11 '16 at 15:27

Denote by $c_1$ the carry-over from the rightmost column, by $c_2$ the carry-over from the next column, by $c_3$ the carry-over from the middle column, and by $c_4$ the carry-over from the fourth column.

Since all four summands and the sum are even, the equation $2N+X+E=10c_1+E$ must be solved with $N,X,E$ even. By distinguishing five cases on $N$, we get the following five possibilities:

       N  X  c1
------------
0  0  0     (Case 0)
2  6  1     (Case 1)
4  2  1     (Case 2)
6  8  2     (Case 3)
8  4  2     (Case 4)


Case 0 has $N=X$, and therefore is illegal.
The next column in the summation yields $3U+Z+c_1=10c_2+Z$, and hence $U=(10c_2-c_1)/3$. Since $0\le c_2\le3$, this yields the following:

       N  X  c1 U  c2
-----------------
2  6  1  3  1  (Case 1)
4  2  1  3  1  (Case 2)
6  8  2  6  2  (Case 3)
8  4  2  6  2  (Case 4)


Now Case 3 has $N=U$, and is illegal.
The middle column in the summation yields $E+U+c_2=10c_3+I$.

• In Cases 1 and 2, this equation becomes $E+4=10c_3+I$. Hence $I$ is also even. The five subcases $E=0,2,4,6,8$ yield $I=4,6,8,0,2$. Since $E$ and $I$ must be different from $N$ and $X$, this leaves only the following subcases alive:

   N  X  c1 U  c2 E  I  c3
--------------------------
2  6  1  3  1  0  4  0   (Case 1a)
2  6  1  3  1  4  8  1   (Case 1b)
4  2  1  3  1  6  0  1   (Case 2 )

• In Case 4, the equation becomes $E+8=10c_3+I$. Hence $I$ is also even. The five subcases $E=0,2,4,6,8$ yield $I=8,0,2,4,6$. Since $E$ and $I$ must be different from $N,X,U$, this leaves only the following subcase alive:

   N  X  c1 U  c2 E  I  c3
--------------------------
8  4  2  6  2  2  0  1   (Case 4 )


Now let us turn to the fourth column, which yields $D+O+c_3=10c_4+E$. Since $E$ is even, this means that $D+O+c_3$ is even.

• In Case 4, the five even digits 0,2,4,6,8 ahave been assigned to $I,E,X,U,N$, respectively. Hence $D$ and $O$ both must be odd, which together with $c_3=1$ yields a contradiction.
• Similarly in Case 2, the only unassigned even digit is $8$. In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4+5$, which forces $c_4=1$ and $\{D,O\}=\{7,8\}$.
• Similarly in Case 1a, the only unassigned even digit is $8$. In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4$, which forces $c_4=1$ and $D+O=10$ with $O$ and $D$ odd and distinct.
• Finally in Case 1b, the only unassigned even digit is $0$. In this case the equation $D+O+c_3=10c_4+E$ becomes $D+O=10c_4+3$, which forces $c_4=1$ and $\{D,O\}=\{0,3\}$.

In all surviving cases, we have $U=3$ and $c_4=1$. In particular, this implies $S=D+1$, so that one of $S$ and $D$ is even.

• In Case 1a, we conclude $S=8$, $D=7$ and $O=3$; this subcase has $U=O$ and yields a contradiction.
• In Case 1b, we conclude $D=0$ and $S=1$, which has $D=0$ as starting digit of a number and hence violates one of the basic principles of alphametic puzzles.
• In Case 2, we finally conclude $D=8$ and $S=9$, and $O=7$.

Let us summarize:

       N  X  c1 U  c2 E  I  c3 D  S  O
-----------------------------------
4  2  1  3  1  6  0  1  8  9  7   (Case 2 )


Hence $I=0$, $X=2$, $U=3$, $N=4$, $E=6$, $O=7$, $D=8$ and $S=9$ have been assigned, and only the digits $1$ and $5$ remain open, and both can be legally assigned to $Z$.

This yields the following two solutions (that only differ in the value of $Z$):

     34               34
34               34
8632     and     8632
+ 87316          + 87356
------           ------
96016            96056

• Hah! you beat me by 2 minutes. – Trenin Feb 11 '16 at 15:27
• But the typing took me almost one hour... – Gamow Feb 11 '16 at 15:28
• I like your notation better. The tables make it very clear. My answer gets bogged down with "If this then .... contradiction. Thus, that...." – Trenin Feb 11 '16 at 15:30

Denote the following:

• $S_i$ to be the sum of the digits in column $i$.
• $D_i$ to be the digit in column $i$ of the sum. Thus $D_i = S_i \mod 10$
• $C_i$ to be the carry over of $S_i$.

Given that all the numbers are even, we know that $N,X,E \in \{0,2,4,6,8\}$. Looking at $S_1=N+N+X+E \le 8+8+6+4=26 \implies C_1 \le 2$.

Lets look at $S_2=U+U+U+Z+C_1$ and $D_2=Z$. We know that $3\times U + C_1 \mod 10 = 0$ in order to make this work. If $C_1=0 \implies U=0$, which can't happen. $C_1=1 \implies U=3$ and $C_1=2 \implies U=6$.

We can also see that $C_4 \le 1$ and since $S \ne D$, we have $C_4=1$ and $S=D+1$.

Now look at $S_1=N+N+X+E$ and $D_1=E$. This implies that $N+N+X \mod 10 = 0$. If $N=0$ then $X=0$, which can't happen.

• $N=2 \implies X=6$ and $C_1=1$ and $U=3$ and $C_2=1$
• $N=4 \implies X=2$ and $C_1=1$ and $U=3$ and $C_2=1$
• $N=6 \implies X=8$ and $C_1=2$ and $U=6$ which can't happen
• $N=8 \implies X=4$ and $C_1=2$ and $U=6$ and $C_2=2$

We also know that $C_3 \le 1$ and $C_4=1$ and $S=D+1$

### Let assume $U=6$.

Thus, $N=8$, $X=4$, and $C_1=C_2=2$.

Then $S_3=E+6+2$. The valid choices for $E \in \{2,0\}$ If $E=0$ then $D_3=I=8$ which is already used. Thus $E=2$, and $I=0$ and $C_3=1$

In $S_4=D+O+C_3=D+O+1$ and $D_4=E=2$. Thus, $D+O=11$. But all the even numbers are already taken, so there is no way to make this equality work.

Thus $U \ne 6$.

### Thus $U=3$

So we know that $N \in \{2,4\}$, $X \in \{6,2\}$, and $C_1=C_2=1$. In either case, $N$ or $X$ is $2$, so no other letter can be $2$. So $E \in \{0,4,6,8\}$.

If $E=8$, then $C_3=1$ and $S_4=D+O+C_3=D+O+1$ and $D_4=E$. This would require one of $D$ or $O$ to be 9, an the other 8.

If $E=4$, then $N=2$ and $X=6$. In $S_3=E+U+C_2=4+3+1=8$, so $D_3=I=8$ and $C_3=0$. From $S_4=D+O+C_3=D+O$ and $D_4=E=4$. The only combination that works are $D+O=9+5$. But either solution for $D$ results in an invalid $S$ since $S=D+1$.

If $E=0$, then from $S_3=E+U+C_2=0+3+1=4$ means that $D_3=I=4$ and $C_3=0$. Thus, $N=2$ and $X=6$. From $S_4=D+O+C_3=D+O$ and $D_4=E=0$. The only combinations of remaining numbers that work for $D$ and $O$ are $9$ and $1$. But neither result in a valid $S$ since $S=D+1$.

Thus $E=6$, then means that one of $D$ or $O$ is 8, and the other is 7. Since $S=D+1$, if $D=7$ then both $N=O=8$. Thus, $D=8$, $O=7$, and $S=9$. Also, $X=2$ and $N=4$. Also, from $S_3=E+U+C_2=6+3+1=10$, so $D_3=I=0$

The final solution is:

$$U=3, E=6, D=8, O=7, S=9, X=2, N=4, I=0, Z\in\{1,5\}$$

in order to be U+U+U+Z = Z , U+U+U needs to be 10 or 20, considering it will get a remainder of 1 or 2 from first line. so only alternatives are 3 or 6. I try giving 6.

        N + N + X = 10 or 20
118
226 x
334
442
550
668 x
776 x
884
992 - will try this

69
69
DE62
+ DO6ZE
-------
SEIZE


I give 3 to E for making min +2 remainder

             69
69
D362
+ DO6Z3
-------
S3IZ3


Here is what I've got now:

N=9, X=2 , U=6 , E=3

remaining: 1,4,5,7,8

so I has to 1. By brute force trying 7 to Z, I get:

             69
69
D362
+ DO673
-------
S3173


N=9, X=2 , U=6 , E=3 , I=1 , Z=7

remaining: 4,5,8

So we have no alternative leaving to get the solution of:

             69
69
4362
+ 48673
-------
53173


because D must be S-1

• The numbers must be even - this restriction was added after your answer. So $N\ne9$. – Trenin Feb 11 '16 at 15:29