# Square dance alphametic

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

     SQUARE
+ DANCE
----------
DANCER


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

• 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... – Oray Mar 9 '16 at 14:34
• @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! – Haobin Mar 9 '16 at 15:33
• The SQUARE DANCE alphametic is not a standard alphametic; the alphametic lovers will see this in one of my later puzzles. – Haobin Mar 9 '16 at 15:33
• @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. – Haobin Mar 9 '16 at 15:35
• @Oray You can add ignore tags on SE. – Carl Löndahl Mar 9 '16 at 16:40

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

Reasoning

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.

Observations

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

SQUARE + DANCE = 10*DANCE + R

and so

SQUARE = 9*DANCE + R

which should lead to some nice properties.

• This doesn't prove that other values of E can't lead to alternate solutions. – Trenin Mar 10 '16 at 18:51
• You're right, it doesn't. I just didn't want to take the time to type it all out. – hexomino Mar 10 '16 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:

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

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