Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.

    + DANCE

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

  • 2
    $\begingroup$ i start to hate alphametrics... tkcs-collins.com/truman/alphamet/alpha_solve.shtml you can even generate your own alphametrics here and u always ask alphametrics... $\endgroup$
    – Oray
    Mar 9, 2016 at 14:34
  • 4
    $\begingroup$ @Oray: If you hate alphametics, then you should not look at puzzles that are clearly tagged [alphametic]. There are many other puzzles on this web site with other tags. Go and look at them! $\endgroup$
    – Haobin
    Mar 9, 2016 at 15:33
  • $\begingroup$ The SQUARE DANCE alphametic is not a standard alphametic; the alphametic lovers will see this in one of my later puzzles. $\endgroup$
    – Haobin
    Mar 9, 2016 at 15:33
  • 2
    $\begingroup$ @Oray: It is also very easy to generate riddles, and there are over 1000 riddles on this web site, and nobody cares whether you would hate them. $\endgroup$
    – Haobin
    Mar 9, 2016 at 15:35
  • 2
    $\begingroup$ @Oray You can add ignore tags on SE. $\endgroup$ Mar 9, 2016 at 16:40

2 Answers 2


Without having checked all possible cases, I've found a solution.

E=3, R=6, C=7, A=1, N=5, U=4, Q=2, D=9, S=8


A useful observation is that we must have S=D-1 and neither can be 0.
Then it seems as though when you pick E, your hand is forced on the other letters.

For example, if we say E=3 then we immediately have to have R=6 and C=7.
Then A+N+1 ends in a 7, but with the digits available this means than the combination A+N must be 1+5, 5+1, 4+2, 2+4.
The next step, U+A=N forces U to be 4, 6, 8 or 2 respectively.
6 is already taken as would be 2 and if we pick the 8 option then we have left no consecutive pair of non-zero digits for S and D. Therefore, we can only have A=1, N=5, U=4.

This means we must have S=8, D=9 as they are the only pair of non-zero consecutive digits left and to make this work we are forced to make Q=2.


There may be an easier way to solve this as we notice


and so


which should lead to some nice properties.

  • $\begingroup$ This doesn't prove that other values of E can't lead to alternate solutions. $\endgroup$
    – Trenin
    Mar 10, 2016 at 18:51
  • $\begingroup$ You're right, it doesn't. I just didn't want to take the time to type it all out. $\endgroup$
    – hexomino
    Mar 10, 2016 at 19:44

To do this exhaustively was painful. Perhaps there is a better way to eliminate some possibilities earlier?

We will note that $C_i$ is the carry over of column $i$.

We have the following equations:

  • $S+C_5 = D \implies C_5=1, D=S+1$ since otherwise, $S=D$ would be a contradiction
  • $Q+D+C_4 =10+A$ since it must generate carry over $C_5=1$
  • $U+A+C_3 \in \{N, 10+N\}$
  • $A+N+C_2 \in \{C, 10+C\}$
  • $R+C+C_1 \in \{E, 10+E\}$
  • $E+E \in \{R, 10+R\}$

Lets work with values of $E$.

If $E=1, R=2, C_1=0$, then

  • $R+C+C_1 = 2+C = 10+E = 11 \implies C=9,C_2=1$.
  • $A+N+C_2=A+N+1=9 \implies A+N=8, C_3=0$.
  • So $A \in \{0,3,5,8\}, N\in\{8,5,3,0\}$.
    • If $A=0, N=8$, then $U+A+C_3=U=N$
    • If $A=3, N=5$, then $U+A+C_3=U+3=5 \implies U=2$
    • If $A=8, N=0$, then $U+A+C_3=U+8=10 \implies U=2$
    • If $A=5, N=3$, then $U+A+C_3=U+5=13 \implies U=8 and C_4=1$
      • Thus $Q+D+C_4=Q+9+1=10+Q \implies Q=A$

If $E=2, R=4, C_1=0$, then

  • $R+C+C_1=4+C=10+E=12 \implies C=8,C_2=1$.
  • $A+N+C_2=A+N+1 \in \{8,18\} \implies A+N \in \{7, 17\}$. 8 is taken so $A+N=7, C_3=0$.
  • If $A,N \in\{1,6\}$, then there are no options for $D=S+1$, so $A,N \in \{0,7\}$ and $S=5, D=6$.
    • If $A=0, N=7$, then $U+A+C_3=U=N$
    • If $A=7, N=0$, then $U+A+C_3=U+7=10+N=10 \implies U=3, C_4=1$.
      • $Q+D+C_4=Q+6+1=Q+7=7 \implies Q=0$

If $E=4, R=8, C_1=0$, then

  • $R+C+C_1=8+C=10+E=14 \implies C=6,C_2=1$.
  • $A+N+C_2=A+N+1 \in \{6,16\} \implies A+N \in \{5, 15\}$. Thus, $A+N=5$.
  • If $A,N \in \{2,3\}$, then there are no options for $D=S+1$, so $A,N \in \{0,5\}, C_3=0$.
    • If $A=0, N=5$, then $U+A+C_3=U=N$
    • If $A=5, N=0$, then $U+A+C_3=U+5=10+N=10 \implies U=A$

If $E=5, R=0, C_1=1$, then

  • $R+C+C_1=0+C+1=E=5 \implies C=4,C_2=0$.
  • $A+N+C_2=A+N \in \{4,14\}$.
  • So $A \in \{1,3,6,7\}, B \in \{3,1,7,6\}$
    • If $A=1, N=3, C_3=0$, so $U+A+C_3=U+1=N=3 \implies U=2, C_4=0$. But $Q+D+C_4=Q+D=11$ leaves no possible values for $Q,D$.
    • If $A=3, N=1, C_3=0$, so $U+A+C_3=U+3=10+N=11 \implies U=8, C_4=1$. But $Q+D+C_4=Q+D+1=10+A=13$ leaves no solution for $Q+D=12$

If $E=6, R=2, C_1=1$, then

  • $R+C+C_1=2+C+1=E=6 \implies C=3,C_2=0$.
  • $A+N+C_2=A+N \in \{3,13\} \implies A+N=13, C_3=1$.
  • If $A,N \in \{5,8\}$, then there are no options for $D=S+1$, so $A,N \in \{4,9\}$ and $D=8, S=7$.
    • If $A=4, N=9$, then $U+A+C_3=U+4=N=9 \implies U=5$
    • If $A=9, N=4$, then $U+A+C_3=U+9=10+N=14 \implies U=5$
  • $Q+D+C_4 \in \{4, 14\}$ has no solution with the remaining digits $\{0,1,5\}$

If $E=7, R=4, C_1=1$, then

  • $R+C+C_1=4+C+1=E=7 \implies C=2,C_2=0$.
  • $A+N+C_2=A+N \in \{2,12\} \implies A+N=12, C_3=1$
  • $A,C\in\{3,9\}$ and $S=5, D=6$.
    • If $A=3, N=9$, then $U+A+C_3=U+3+1=U+4=9 \implies U=5$
    • If $A=9, N=3$, then $U+A+C_3=U+9+1=10+N \implies U=N$

If $E=8, R=6, C_1=1$, then

  • $R+C+C_1=6+C+1=E=8 \implies C=1,C_2=0$.
  • $A+N+C_2=A+N \in \{1,11\} \implies A+N=11, C_3=1$.
  • Thus, $A\in \{2,9,4,7\}, N\in\{9,2,7,4\}$.
    • If $A=2, N=9$, then $U+A+C_3=U+2+1=N=9 \implies U=6$
    • If $A=9, N=2$, then $U+A+C_3=U+9+1=10+N \implies U=N$
    • If $A=4, N=7$, then $U+A+C_3=U+4+1=N=7 \implies U=2$, which leaves no solution for $D=S+1$.
    • If $A=7, N=4$, then $U+A+C_3=U+7+1=20+N=24 \implies U=6$

If $E=9, R=8, C_1=1$, then

  • $R+C+C_1=8+C+1=E=9 \implies C=0,C_2=0$.
  • $A+N+C_2=A+N=10, C_3=1$.
  • So $A\in\{3,7,4,6\}, N\in\{7,3,6,4\}$.
    • If $A=3, N=7$, then $U+A+C_3=U+3+1=U+4=7 \implies U=3$.
    • If $A=7, N=3$, then $U+A+C_3=U+7+1=U+8=13 \implies U=5, C_4=1$. This leaves $S=1, D=2$.
      • $Q+D+C_4=Q+2+1=Q+3=10+A=17 \implies Q=14$
    • If $A=4, N=6$, then $U+A+C_3=U+4+1=U+5=6 \implies U=1, C_4=0$. This leave $S=2, D=3$.
      • $Q+D+C_4=Q+3=10+A=14 \implies Q=11$
    • If $A=3, N=7$, then $U+A+C_3=U+3+1=U+4=7 \implies U=3$.

Thus, $E=3, R=6, C_1=0$.

We can also know that $R+C+C_1=6+C=10+E=13 \implies C=7, C_2=1$.

Also, $A+N+C_2=A+N+1 \in \{7,17\}$. Thus, $A+N \in \{6,16\}$. But 16 is not possible with 7 already taken, so $A+N=6, C_3=0$.

We know that $A\in \{1,5,2,4\}, N\in\{5,1,4,2\}$. Either way, the only solution for $D=S+1$ is $S=8, D=9$

  • If $A=5, N=1$, then $U+A+C_3=U+5=10+N=11 \implies U=6$.
  • If $A=2, N=4$, then $U+A+C_3=U+2=N=4 \implies U=2$
  • If $A=4, N=2$, then $U+A+C_3=U+4=10+N=12 \implies U=8$.

Thus, $A=1, N=5$. Therefore $U+A+C_3=U+1=N=5 \implies U=4, C_4=0$. Then $Q+D+C_4=Q+9=10+A \implies Q=2$. Thus, the solution is:

$$E=3, R=6, C=7, S=8, D=9, A=1, N=5, U=4, Q=2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.