Acids VS Bases

Let the right-most column be column $0$ .

• From column 0, we note that: $(0): \quad E+A = \begin{cases} S \\ S + 10 \end{cases}$ . Let's denote it as $E + A = S + ^0_{10}$ . And column 1: $(1): \quad 2S + ^0_1 = E + ^0_{10}$ . Therefore, $(0)+(1): \begin{cases} E + A + 2S = S + E + ^0_{10} & \Rightarrow A + S = ^0_{10} > 0 & \Rightarrow A+S = 10 \\ E + A + 2S + 1 = S + 10+ E + ^0_{10} & \Rightarrow A + S = ^{-1}_{19} & \Rightarrow \color{#D61F06}{\text{No solution}} \end{cases}\\ \Rightarrow \color{#3D99F6}{(0): \quad E+A=S} \quad \Rightarrow \color{#3D99F6}{A+S = 10}$

• From $(0): \quad E+A = S \quad \Rightarrow E = S - A = 10 - 2A \quad \Rightarrow S > E, S > A, A = 1,2,3,4$ (denoted by: $\color{#3D99F6}{A^1_4} : A = 1$ to $4$ ) $\quad \Rightarrow \color{#3D99F6}{E^{10-2(1)}_{10-2(4)}=E^8_2}$ $\quad \Rightarrow \color{#3D99F6}{S^{10-1}_{10-4}=S^9_6}$

• From $(2): \quad A^1_4 + B + ^0_1 = S^9_6 + ^0_{10} \quad \Rightarrow B + ^0_1 = ^8_2 + ^0_{10} \quad \Rightarrow \color{#3D99F6}{B + A +^0_1 = S} \quad$ Since $(0): \quad E+A = S$ $\Rightarrow B+ 1 = E \quad \Rightarrow \color{#3D99F6}{B^7_1}$

• Since $(2): \quad B + A + 1 = S \quad \Rightarrow (3): \quad 2L + 0 = A^1_4 \quad \Rightarrow \color{#3D99F6}{L^1_2} \quad \Rightarrow \color{#3D99F6}{A=2, 4}$ but if $L = 2$ , then $E = 10 -2A = 2 = L$ which is $\color{#D61F06}{unacceptable}$ . Therefore, $\color{#3D99F6}{L=1, A = 2, E = 6, S = 8, B = 5}$

$\Rightarrow B+A+S+E+S = 5+2+8+6+8 = \boxed{29}$

There's no change in the ten thousands, so L < 5. Let's look at hundreds and ones, one of the summands is different although they have the same sum. This must be caused by the carry forward from the tens, so B = E - 1. Since we've established the existence of the carry forward from tens, then S ≥ 5 and E < S, always. By the same reason, A < S and A + E = S < 10, so there won't be any other carry forwards sent over from ones and hundreds (thousands doesn't have one either, but that's because L < 5). E & A is an even number while B is odd. Since E & A as summands are both even, then S as their sum must also be even.

There are two ways for E + A = S to happen, one is 2 + 4 = 6 and the other 6 + 2 = 8, but the first solution will have L = 2 = E, which shouldn't be for they represent distinct digits each.

Finally, we have E = 6, A = 2, L = 1, S = 8 and B = 5.

Answer = B + A + S + E + S
= 5 + 2 + 8 + 6 + 8
= 29

I use a different approach. From the column (1), we know, that if we call S = E+A, that 2(E+A) mod 10 = E. The integer solution of this equation give four some possible subset {A, E} = {4,2}, {2,6}, {3,4}, {8,1}. In particular the first subset cannot be accepted because from the column (4) we know that 2L = A, so L=2. The last information that help us to chose the correct subset came from the column (3). In fact from the column we know that E+A = A+B, but it is only possible in two case: when B = E, but is impossible for hypothesis, and when B = E-1, that is possible only if S must be ≥ 5 with carry-over of 1. This information automatically exclude the subset {3,4} because yet A is set to 3. So remain {2, 6} and {8, 1} but if we want respect the (2) column we must impose that A < E, that automatically exclude {8, 1}. So the correct association is {A, E} = {2, 6}, that means {L, B, S} = {1, 5, 8}