jmc

We must solve the addition problem \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c} & & S & E & A & S_d \\ & & & E & B & B_d \\ + & & & S & E & A_d\\ \cline{1-6} & & B & A& S& S_d\end{array}, where $d$ is an unknown base. It follows that $S+B+A$ leaves a residue of $S$ upon division by $d$. Thus, $B+A$ must be divisible by $d$. Since $B$ and $A$ cannot both be $0$, and $B+A<(d-1)+(d-1)=2d-2$, then $B+A=d$.

Looking at the $d$s digit, we must carry-over $1$ from the units digit sum, so $1+A+B+E\equiv S(\mathrm{mod}d)$. Since $B+A=d$, then $1+E+d\equiv 1+E\equiv S(\mathrm{mod}d)$. Thus, $S=E+1$ or $E=d-1$ and $S=0$. However, the latter is impossible since $S$ is the leftmost digit of 'SEAS' and 'SEA'. Thus, $S=E+1$, and we again carry-over $1$ to the $d^2$ digit.

Looking at the $d^2$ digit, after carry-over, it follows that $1+E+E+S\equiv A(\mathrm{mod}d)$. Note that $1+E+E+S<1+3(d-1)=3d-2<3d$. Then, $2E+S+1-A$ is either equal to $0$, $d$, or $2d$. However, we can immediately discard the $0$ case: there would be no carry-over for the leftmost digit, so $S=B$ are not distinct.

In the next case, if $2E+S+1=A+d$, then there is a carry-over of $1$ to the last digit. It follows that $S+1=B$. This gives us the system of equations $$\begin{array}{rl}B+A& =d\\ E+1& =S\\ S+1& =B\\ 2E+S+1-A& =d\end{array}$$ Setting the first and fourth equations equal to each other yields that $d=B+A=2E+S+1-A$, and since $B=S+1=E+2$, substituting for $B$ and $S$ yields that $2A=3E+S+1-B=2E+(E+1)+1-(E+2)=2E$. This contradicts the distinct digits criterion.

Thus, $2E+S+1-A=2d=2(B+A)$, so $2E+S+1-2B=3A$. Also, we have that $B=S+2$, due to the carry-over in the leftmost digit. Substituting for $B$ and $S$ yields that $3A=2E+(E+1)+1-2(E+3)=E-4$, so $E=3A+4$. Thus, $S=3A+5$ and $B=3A+7$. Also, $S,E,$ and $A$ are decimal digits, so it follows that $S=3A+5\le 9⟹A=0,1$. We can discard the solution $A=0$, as $d=B+A$ but $B<d$. Thus, $B=10,S=8,E=7$, occurring in base $d=B+A=11$. The answer is $\boxed{871}$.