Single cryptogram, quadruple solutions?

$\begin{array}{c}&&&C&H&A&R \\ +&&&R&A&Y\\ +& S&A&U&R&A\\ \hline & E&M&B&E&R \end{array}$

Terminology : I read the cryptogram from left to right, so I will refer to the leftmost column as the first column, and so on..

First, note that if we can find one set of solution, we can easily find another by just interchanging the numbers representing $H$ and $U$ in the cryptogram. Doing so will not affect other parts of the cryptogram, since we are interchanging numbers in a single column which do not appear elsewhere in the cryptogram.

This explains why Ray is so excited at the beginning. He thought the two sets of solutions just have their $H$ and $U$ interchanged, proceeded to divide $1256$ by $2$ to get $\overline{RAY}=628$ . However, he soon discovered that this is not the case. Note that if $Y=8, R=6, A=2$ , we can deduce that $E=1$ by looking at the last two columns. This is a contradiction, since if $E=1$ , then from the first column, $S<1$ , which leaves only $1$ option, that is $S=0$ , but we (and Ray) know full well that this cannot be.

Therefore, Ray concludes once he found the two sets of solutions which match the conditions set by Scythe, he can easily interchange the $H$ and $U$ in both sets of solutions to get $2$ more solutions.

Now, we move on to the task of finding the solutions. First, every congruence you see will be in modulo 10. We will not specify this in every congruence since there are a large number of them. Note that from the last column $R+Y+A \equiv R$ . This can only happen if $Y+A \equiv 0$ . Since $Y+A<20$ , we conclude that $Y+A=10$ and there is a carry over of $1$ from the last column to the fourth column. From the fourth column, this implies that $2A+R+1 \equiv E$ .

From the first column, we can conclude that $S+1=E$ with a carry over of $1$ from the second column. Why a carry over of $2$ is not possible is left to the reader.

From the hint that the two $\overline{RAY}$ sums up to $1256$ , we can write this congruence: $Y_1+Y_2 \equiv 6$ , where $Y_1$ and $Y_2$ denotes the two solutions of $Y$ . We will now investigate the possible values of $Y$ .

Suppose that $Y=0$ , then from $Y+A=10$ , $A=10$ , a clearly impossible case.

Suppose that $Y_1=1$ , then we conclude that $Y_2=5$ and as a consequence $A_2=5$ , again, clearly impossible since $Y_2$ and $A_2$ must be different numbers.

For $Y_1=2, A_1=8$ , note that $Y_2=4, A_2=6$ . As a consequence, $\overline{R_{1}82} + \overline{R_{2}64} = 1256$ , or $(R_1+R_2)(100)=1110$ , clearly a contradiction.

Now, we move on to the case where $Y=3, A=7$ . Note that the congruence $Y_1+Y_2 \equiv 6$ forces $Y_1, Y_2=3$ and $A_1,A_2=7$ . We thus obtain $(R_1+R_2)(100)+(73)(2)=1256$ , or $(R_1+R_2)(100)= 1110$ , an impossibility.

Now, suppose that $Y_1=4$ . This would mean that $Y_2=2$ , but we have already proven that $Y \neq 2$ !

Suppose that $Y=5$ , then note that $A=5$ , an impossibility. If $Y_1=6$ , then $Y_2=0$ , a clear impossibility as we have proven above.

If $Y_1=8$ , then $Y_2=8$ , and the two $A$ s= $2$ . As a consequence, $(R_1+R_2)(100)+(28)(2)=1256$ or $R_1+R_2=12$ . Possible combinations for the two $R$ s are $(3,9),(4,8),(5,7),(6,6)$ . The cases of $(9,3),(8,4),(7,5)$ are symmetrical.

We can straightaway rule out $(4,8)$ since already $Y=8$ . The case of $(6,6)$ has been discussed at the very beginning, and therefore is ruled out as well. From the congruences $2A+R+1 \equiv E$ and $S+1 \equiv E$ , we obtain $2A+R \equiv S$ . Now, let's suppose that $R_1=3$ . From the congruence $2A+R \equiv S$ , we can deduce that $S=7$ and $E=8$ , but this is an impossibility since already, $Y=8$ . Suppose now that $R_1=5$ , then from the congruence again, we get $S=9$ and $E=0$ , clearly an impossibility.

Therefore, all that is left for possible values of $Y$ and $A$ are $Y_1=7, A_1=3$ and $Y_2=9, A_2=1$ . Now, finally that is some progress!

Now, note that $\overline{R_{1}37} + \overline{R_{2}19} = 1256$ , or $R_1+R_2=12$ . The possible values for $(R_1,R_2)$ in ordered pairs are $(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)$ . We can immediately spot the deal breakers $(3,9)$ and $(7,5)$ . (Why?)

To eliminate the option $(4,8)$ , we focus on the values of $R_2$ and $A_2$ . Note that $R_2=8$ and $A_2=1$ , and from the congruence $2A+R \equiv S$ , $S=0$ , which we know is impossible.

To eliminate the option $(5,7)$ , note that $R_2=7$ and $A_2=1$ forces $S_2=9$ . However, $9$ is already represented by $Y_2$ .

To eliminate the option $(6,6)$ , note that $R_2=6$ and $A_2=1$ forces $S_2=8$ and $E_2=9$ . However, this is a contradiction since $9$ is already represented by $Y_2$ .

Now, we are left with $2$ options of $(Y_1,A_1,E_1,S_1,R_1,Y_2,A_2,E_2,S_2,R_2)$ , namely $(7,3,6,5,9,9,1,6,5,3)$ and $(7,3,5,4,8,9,1,7,6,4)$ . We will prove that the first option is wrong.

For the first option, let's focus on the values of $(Y_1,A_1,E_1,S_1,R_1)$ and attempt to complete the cryptogram.

$\begin{array}{c}&&C&H&3&9 \\ +\quad &&9&3&7\\ + \quad 5&3&U&9&3\\ \hline \qquad 6&M&B&6&9 \end{array}$

Now, the available digits are $0,1,2,4,8$ . Note that there is a carry over of $1$ from the second column to the first column. In order for this to be possible $C$ must equal $8$ . (Why?)

Now, notice that there is a carry over of $1$ from the fourth column to the third column. This implies that $H+U \equiv B$ , which we must satisfy using $0,1,2,4$ . Obviously, this is impossible. Therefore, we reject the first option and now are left only with the second option.

For the second option, let's focus on the values of $(Y_1,A_1,E_1,S_1,R_1)$ and attempt to complete the cryptogram first.

$\begin{array}{c}&&C&H&3&8 \\ +\quad &&8&3&7\\ + \quad 4&3&U&8&3\\ \hline \qquad 5&M&B&5&8 \end{array}$

The unused digits are $0,1,2,6,9$ . Focus on the second column. In order for there to be a carry over to the first column, note that $C=6$ or $C=9$ .

Suppose that $C=9$ , then $M=2$ . There cannot be any carry over from the third column (Why? Think about the values M can take!). Notice that there is a carry over of $1$ from the fourth column to the third column, implying the equality $9+H+U=B$ (no carry over, remember!). However, it is now pretty obvious that this is impossible.

Therefore, $C=6$ . If we attempt to complete the congruence $9+H+U \equiv B$ using only the values $0,1,2,9$ , the only possible combinations are $(H,U,B)=(0,2,1),(2,9,0)$ and another pair where $H$ and $U$ interchange. For the first option, note that $M=9$ . This cannot be true since there is a carry over of $1$ from the third column. For the second option, note that $M=1$ . Checking, we find that everything holds true. We have thus find the first two solutions to the cryptogram!

Now, we proceed to find the second pair of solutions. The partially completed cryptogram with values of $(Y,A,E,S,R)=(9,1,7,6,4)$ is as below:

$\begin{array}{c}&&C&H&1&4 \\ +\quad &&4&1&9\\ + \quad 6&1&U&4&1\\ \hline \qquad 7&M&B&7&4 \end{array}$

The unused digits are $0,2,3,5,8$ . Note that in order for a carry over of $1$ from the second column to the first column to be possible, $C=8$ . Now we attempt to complete the congruence $4+H+U \equiv B$ using the values of $0,2,3,5$ . We find only two possibilities, that is $(H,U,B)=(5,3,2), (3,5,2)$ . This also implies that $M=0$ . Checking the solution, we find that everything holds true. Thus, we have our second pair of solutions.

A quick computation reveals the the sum of all $4$ possible values of $\overline{CHAR}$ is $8314+8514+6238+6938=\boxed{30004}$ .

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On a totally unrelated note, if anyone manages to find extra solutions to the cryptogram above (without any conditions attached), please let me know. I have zero programming knowledge and am too unmotivated to work through the large number of cases by hand.