0
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The following puzzle has a spelling mistake: DOUCE should be DOUZE.
The correct version is: French alphametic (corrected)
(Surprisingly the faulty version allows feasible solutions.)

Every letter stands for a digit in base-10 representation, different letters stand for different digits:

         UN 
         UN
       DEUX
   +  DOUCE
  ------------
      SEIZE

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

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  • $\begingroup$ I am not sure this can be solved without a programming. $\endgroup$ – Oray Feb 10 '16 at 13:50
  • 2
    $\begingroup$ There are thirteen solutions to this puzzle. I don't think finding them all without the help of a computer would be a very rewarding task... $\endgroup$ – squeamish ossifrage Feb 10 '16 at 15:15
  • 6
    $\begingroup$ Hello, French citizen here. Just to point out that "DOUCE" is not a number. Not that it hinders the puzzle in any way, though. For the trivia part, "douze" is French for twelve, but "douce" is the feminine version of "doux", which means soft/sweet. $\endgroup$ – Chester Copperpot Feb 10 '16 at 17:23
  • $\begingroup$ @ChesterCopperpot And with DOUZE there are twelve solutions, which sort of fits I guess :-) $\endgroup$ – squeamish ossifrage Feb 10 '16 at 18:41
  • $\begingroup$ Stupid me, I made a mistake in typing, DOUCE should be DOUZE so that 1+1+2+12=16. I'll upvote and accept the answer below, and post a corrected version as a separate puzzle. I am very sorry for my mistake. $\endgroup$ – Haobin Feb 11 '16 at 13:16
7
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sorry for the format I don't know how to put spoiler to multiple lines of formula. I am just putting spoiler to the answer below.

remaining numbers : 0-1-2-3-4-5-6-7-8-9

             UN
             UN
           DEUX
        + DOUCE
        -------
          SEIZE

Giving 0 to E , and since we know S = D + 1 I random them as 8 - 9. I will try giving 2 to O . This is what we have:

E=0 , D=8 , S=9, O=2, remaining numbers : 1-3-4-5-6-7

              UN
              UN
            80UX
        +  82UC0
         -------
           90IZ0

Now we know that :

N+N+X = 10 or 20

U+U+U+C + B = Z where B = 1 or 2

U + 0 = I but we know they can't be the same so there is a remainder which means:

U + W = I where W = 1 or 2

        N N X possibilities:
        a1: 7+7+6
        a2: 3+3+4
        a3: 2+2+6

To determine whether if W or B can be 1 or 2 I pick a2 to give big numbers to net stage and I get;

E=0 , D=8 , S=9, O=2, N=3, X=4, remaining numbers : 1-5-6-7

             U3
             U3
           80U4
        + 82UC0
        -------
          90IZ0

U+U+U+C + 1 = Z

U + W = I where W = 1 or 2

U can be 5 or 6, and I can 6 or 7 because of remaining numbers.

UUUC possibilities:

6661 +1 = 20 x Z can't be 0
6665 +1 = 24 x Z can't be 4
5557 +1 = 23 x Z can't be 3
5551 +1 = 17 so let's give our values and rewrite:

ANSWER:

E=0 , D=8 , S=9, O=2, N=3, X=4, U=5, C=1, Z=7, I=6

             53
             53
           8054
        + 82510
        -------
          90670

Is my solution but I am sure there are more possibilities and this is not a unique solution because of many possibilities.

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  • $\begingroup$ Why are we assuming E = 0? $\endgroup$ – question_asker Feb 10 '16 at 14:53
  • $\begingroup$ add ">!" at the beginning of each line to hide multi-line spoilers. $\endgroup$ – friedemann_bach Feb 10 '16 at 14:56
  • $\begingroup$ Not we, it is my solution both containing brute force and logic. I gave the working solution, there were dead-ends while I was solving. $\endgroup$ – canova Feb 10 '16 at 14:57
  • $\begingroup$ there are actually 10 solutions. $\endgroup$ – Oray Feb 10 '16 at 16:47
3
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There are 13 solutions;

1.

C=3 D=1 E=0 I=5 N=6 O=9 S=2 U=4 X=8 Z=7

2.

C=6 D=1 E=0 I=7 N=8 O=9 S=2 U=5 X=4 Z=3

3.

C=7 D=3 E=0 I=2 N=1 O=6 S=4 U=9 X=8 Z=5

4.

C=1 D=3 E=0 I=6 N=9 O=7 S=4 U=5 X=2 Z=8

5.

C=7 D=5 E=0 I=1 N=9 O=4 S=6 U=8 X=2 Z=3

6.

C=3 D=5 E=4 I=7 N=1 O=9 S=6 U=2 X=8 Z=0

7.

C=3 D=7 E=1 I=9 N=5 O=4 S=8 U=6 X=0 Z=2

8.

C=0 D=7 E=3 I=4 N=9 O=6 S=8 U=1 X=2 Z=5

9.

C=3 D=7 E=4 I=1 N=9 O=6 S=8 U=5 X=2 Z=0

10.

C=1 D=8 E=0 I=6 N=3 O=2 S=9 U=5 X=4 Z=7

11.

C=0 D=8 E=1 I=7 N=4 O=3 S=9 U=5 X=2 Z=6

12.

C=4 D=8 E=2 I=1 N=5 O=3 S=9 U=7 X=0 Z=6

13.

C=0 D=8 E=2 I=3 N=7 O=4 S=9 U=1 X=6 Z=5

found it by using a simple alphametic solver.

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  • $\begingroup$ Seems to me the solution are 15 $\endgroup$ – dmg Feb 11 '16 at 13:51

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