Alphametic twins

Question

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

      LET               SETS
    + THE             + TIRE
   -------           --------
     LASS              LATER

These alphametic twins have a unique solution. This makes them special and interesting. The left alphametic alone has many solutions. The right alphametic alone has many solutions.

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

Answer 1

L = 1
S = 2
A = 0
T = 8
E = 4
R = 6
H = 7
I = 3

Explanation:

First of all, note that L must be 1. This is because the maximum LASS can be is 999 + 999 = 1998, and L cannot be 0. (An equivalent argument can be given for LATER, if you want, instead). Another argument can be that the maximum you can carry over from a column is '1', and column 2 obviously must carry, which means L must be 1.

L = 1

Next, compare the first 2 columns of the twins. They are exactly the same except for the L and S. L and S cannot be equal, however. The only way they can be different is if column 3 (counting columns from left to right) carries in one alphametic but does not in the other. This would mean L and S must have a difference of 1. As S cannot be 0, S must therefore be 2, and it must be the left alphametic which carries in column 3.

S = 2

Now that we know the third column does not carry in the second alphametic, we can know for sure by considering the first 2 columns that: 2 + T = 10 + A ⇒ T = 8 + A
Therefore T is 8 more than A. A and T are single digits, so the only possibilities for A and T are {A = 0, T = 8} or {A = 1, T = 9}. However, L = 1, and no 2 letters are the same digit, so:

A = 0
T = 8

Consider column 4 (units column) in the first alphametic. We know T = 8 and S = 2. Therefore E = 4 and that column must carry.

E = 4

Consider column 5 (units column) in the second alphametic. We know S = 2 and E = 4. Therefore R = 6 and that column does not carry.

R = 6

Consider column 3 in the first alphametic. As stated before, since T = 8 and E = 4, column 4 must carry '1' over to column 3. We know that E = 4 and S = 2. Therefore, this column must also carry since S is smaller than E, and so: E + H + 1 = 10 + S ⇒ 4 + H + 1 = 10 + 2 ⇒ H = 7

H = 7

Column 5 in the second alphametic does not carry, as stated before. Considering column 4, T = 8, R = 6 and E = 4. Therefore column 4 does carry. Considering column 3: E = 4 and T = 8. This column does not carry, so therefore: E + I + 1 = T ⇒ 4 + I + 1 = 8 ⇒ I = 3

I = 3


The solution works as it satisfies all above requirements (no 2 digits are the same, and no leading digits are zero). In addition, all of the above steps of logic were forced, so it must be the unique solution.