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
Commented Feb 11, 2016 at 13:25
• @Oray Now there are 2 solutions - he added an additional restriction that all the numbers are even. Commented Feb 11, 2016 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. Commented Feb 11, 2016 at 15:27
• But the typing took me almost one hour... Commented Feb 11, 2016 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...." Commented Feb 11, 2016 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 $N\ne9$. Commented Feb 11, 2016 at 15:29