# An almost Shakespearian alphametic

Every letter stands for a digit in base-9 representation, different letters stand for different digits, and leading digits are always non-zero.

       ALLS
+  WELL
+  THAT
+   END
---------
SWELL


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

• there are 30 solutions -.-
– Oray
Mar 8, 2016 at 13:56
• By base 9 you mean I can use digits 1 to 8 or 9 is allowed, or I'm writing 9 in base 9 as 10 (which would be strange)? Mar 8, 2016 at 14:45
• @Oray Actually, in base 9, I got 2 solutions. Mar 8, 2016 at 14:47
• it was 10 originally :)
– Oray
Mar 8, 2016 at 14:48
• @Oray. So Am I allowed to use the digit 9 or not. Sorry but It's not clear to me so I'm asking bluntly :) Mar 8, 2016 at 15:01

First, we see that $S\in\{1,2\}$. From the first column, we see that $S+L+T+D$ yields $L$. Thus, $S+T+D\in \{10,20\}$. Since S is so small, we can conclude that $S+T+D=10$.

From the second column, we see $L+L+A+N$ yields $L$. Since there is a carry over of 1, $L+A+N\in\{8,18\}$.

From the third column, we see $L+E+H+E$ yields $E$. Since the carry over is 1 or 2, we know $L+E+H\in\{7,8,17,18\}$.

Lastly, $A+W+T$ yields $W$. Again, carry over is 1 or 2, so $A+T\in\{7,8,17,18\}$. But only $\{7,8\}$ are really possible, so we know that S=1.

Also, we know that $T+D=8 \implies A+T=7$, so we know that $A=D-1$. Also, we know the carry over from the 3rd column must be 2, so $L+E+H\in\{17,18\}$. The possible values are then:

• $A \in \{2,4,5\}$
• $T \in \{5,3,2\}$
• $D \in \{3,5,6\}$

Thus, 5 is not an option for any other letter since it must be used by one of these three.

Lets look at the letter $L\in\{0,2,3,4,6,7,8\}$.

• If $L=0$, then $L+E+H$ cannot be made 17 or 18.
• If $L=2$, then $A+N=6 \implies N=T-1$ which means either $N$ or $A$ must be 2.
• If $L=3$, then $A+N=5 \implies N=T-2 \implies N=0, T=2, A=5, D=6$. But then $E+H=15$ cannot be made since 6 in taken.
• If $L=4$, then $A+N=4 \implies A=N=2$. Thus, $A+N=14 \implies A=5, T=2, D=6, N=8$. Again, this makes $E+H=13$ impossible with the remaining digits.
• If $L=8$, then $A+N=10 \implies N=T+2$. Thus, $A=5, T=2, D=6, N=4$. Again, $E+H=8$ is impossible with the remaining digits.
• If $L=7$, then $A+N=11 \implies N=T+3$. Thus, either $A=2,T=5,D=3,N=8$ or $A=4,T=3,D=5,N=6$. In both cases, $E+H=10$ are impossible.

Thus $L=6$.

So, $A+N=12$ or $A+N=2$. If $A+N=12$, then $N=T+4$. Thus, $A=4, T=3, D=5, N=7$. Also, the carry over is 2 so that $L+E+H=17$. Thus, $E+H=11$ so $E,H \in \{8,2\}$. This leaves $W=0$, but we cannot have a 3 digit number.

Therefore, $A+N=2$. Thus, $A=2, N=0, T=5, D=3$. $L+E+H=18$ so $E+H+12$. This requires $E,H\in\{4,6\}$.

So the valid solutions are:

 2661
8466
5725
+ 403
-----
18466


And

 2661
8766
5425
+ 703
-----
18766


Where:

$$A=2, D=3, L=6, N=0, S=1, T=5, W=8, E,H\in\{4,7\}$$

Solution(s):

A = 2
L = 6
S = 1
W = 8
E = 4/7
T = 5
H = 7/4
N = 0
D = 3

Explanation In order to make it easier my explanation are going to be made in base 10. So if I say A+W = 10 this means in base 10, In base 9 it would be A+W = 11 (which is 10 in base 9). Headache already? :) I will only specify base 9 where is needed

SWELL - WELL = S0000 (in base 9). this means that
ALLS +
THAT +
END =
S0000 (base 9).
S has to be 1. In order for it to be 2, A+T+ carriage from the hundreds should be above 18 (20 in base 9). Since they can be max 8+7 it means the carriage from the tens should be 3 at least. And L+H+E can be max 6+5+4. You cannot add up to 27 (30 in base 9) by adding the carriage from the units position.
ALL1 +
THAT +
END =
10000 (base 9).
This means T+D = 8. Also T+A+carriage from hundreds = 9 (10 in base 9). This means T+A = 7 or 8. It cannot be 8 because this will result in A = D. So T+A = 7.
This results in D = A+1 so D >= 1. But if D = 2 then A = 1 = S so D >= 3. Let's start with D = 3. So A = 2 and T = 5
2LL1 +
5H25 +
EN3 =
10000 (base 9)
this means L + 2 + N + 1 (carriage) = 9 or 18 (base 10). So L + N = 6/15 (base 10).
L + N = 6 can be obtained from the remaining digits only if one is 0 and one is 6.
Let's say L = 6 and N = 0.
2661 +
5H25 +
E03 =
10000 (base 9) this means H + 6 + E + 1 (carriage) = 18 (base 10). H + E = 11 (base 10). Only available values for a sum of 12 (base 10) are 7 and 4.
Works in both combinations. but let's say H = 7 and E = 4 Now to the original sum 2661 +
W466 +
5725 +
403 =
1W466
W can be anything but the only available value is 8. So. Solution 1 2661 +
8466 +
5725 +
403 =
18466 (base 9)
A = 2 L = 6 S = 1 W = 8 E = 4 T = 5 H = 7 N = 0 D = 3 Solution 2 reversing H and E 2661 +
8466 +
5725 +
403 =
18466 (base 9)
A = 2 L = 6 S = 1 W = 8 E = 7 T = 5 H = 4 N = 0 D = 3 When we started this we assumed D = 3. Let's increment. D = 4. From T + D = 8 we get a "No go" since T = D = 4. For D = 5 We get T = 3 and A = 4.
4LL1 +
3H43 +
EH5 =
10000 (base 9) So L + H + 4 + 1 = 9 or 18 (base 10). L + H = 4 or 13. So L and H can be 6 and 7. L = 6 results in 4661 +
3743 +
EN5 =
10000 (base 9).
This results in N = 7 which cannot be since H = 7. same goes for L = 7 and H = 6.
For D = 6 or more we get similar results following the same path.

• Sorry for the formatting. I'm still new to this "spoiler" blockquotes. I'm working on it. Mar 8, 2016 at 16:13