Sherlock Holmes Base Number Problem

Question

Sherlock Holmes and Dr. Watson recover a suitcase with a three-digit combination lock from a mathematician turned criminal. Embedded in the suitcase above the lock is the cryptic message "AT SEA BASE. SEAS EBB SEA: BASS. "

Dr. Watson comments, "This probably isn't about ocean fish. Perhaps it is an encrypted message. The colon suggests the addition problem $SEAS + EBB + SEA = BASS$, where each letter represents a distinct digit, and the word 'BASE' implies that the problem is in a different base."

Holmes calmly turns the combination lock and opens the suitcase. While Dr. Watson gapes in surprise, Holmes replies, "You were indeed right, and the answer was just the value of the word $SEA$, interpreted as decimal digits." What was the lock combination?

I'm finding this problem really hard. I think this is a Mathcounts problem. I have tried setting the base to different numbers and then bashing it but I got nowhere. Help!

Answer 1

Suppose that each letter represents a distinct digit, and that the problem is in a different base, say $x$. Then we can represent the problem as follows: \begin{eqnarray*} Sx^3+&Ex^2+&Ax+&S&\\ &Ex^2+&Bx+&B&\\ &Sx^2+&Ex+&A&\quad+\\ \hline Bx^3+&Ax^2+&Sx+&S& \end{eqnarray*} In each column we have at most $2$ carries, so from this it follows that \begin{eqnarray*} S+B+A &=& S+k_1x,\\ A+B+E+k_1&=& S+k_2x,\\ E+E+S+k_2&=& A+k_3x,\\ S+k_3 &=& B, \end{eqnarray*} where $0\leq k_1,k_2,k_3\leq2$, and $0\leq A,B,E,S<x$ are distinct.

The first equation shows that $B+A=k_1x$, so $k_1=1$ because $0\leq A,B<x$ are distinct, so $$B=x-A\tag{1}.$$ Then in the second equation we see that $k_2\geq1$ because $A+B=x$, and we find that $$E=S+(k_2-1)x-1.$$ Because $0\leq E<x$ we have either $k_2=1$ and $E=S-1$, or $k_2=2$ and $S=0$ and $E=x-1$. In the latter case, the third equation shows that $$A+k_3x=E+E+S+k_2=(x-1)+(x-1)+0+2=2x,$$ and hence $k_3=2$ and $A=0=S$, contradicting the assumption that the digits are distinct. Therefore $k_2=1$ and $$E=S-1.\tag{2}$$ The fourth equation shows that $k_3\neq0$ because $S\neq B$, and the third equation now yields $$A=3S-2-k_3x.$$ Plugging in $A=x-B$ and $S=B-k_3$ and rearranging then yields $$4B+1=(k_3+1)(x+3),$$ which shows that $k_3$ is even, so $k_3=2$. Then \begin{eqnarray*} B&=&\frac{3x+8}{4}&=&\frac34x+2,\\ A&=&x-B&=&\frac14x-2,\\ S&=&B-2&=&\frac34x,\\ E&=&B-3&=&\frac34x-1,\\ \end{eqnarray*} which shows that $x$ is divisible by $4$ and $x\geq12$ because $B<x$. Because $S,E,A<10$ it follows that $x=12$, and $(S,E,A)=(8,7,1)$.

Answer 2

The lock combination was $871$. This can be proved as follows. First, since $S$, $E$ and $A$ have to be interpreted as digits, there are $\leqslant 9$. Let $N$ be the unknown base. One has \begin{matrix} &S&E&A&S\\ +&&E&B&B\\ +&&S&E&A\\ \hline &B&A&S&S \end{matrix} which gives successively

1. $S + B + A = S + N$, whence $A + B = N$ and $A > 0$,
2. $1 + A + B + E = S + N$, whence $S = E + 1$
3. (a) $1 + E + E + S = A + N$ and $B = S + 1$ or (b) $1 + E + E + S = A + 2N$ and $B = S + 2$.

Case (a) would lead to $2 + 3E = 2A + E + 2$, whence $A = E$ and this would not be a code. Case (b) leads to $2 + 3E = 3A + 2E + 6$, whence $E = 3A + 4$. Since $E \leqslant 9$ and $A > 0$, one has necessarily $A = 1$, $E = 7$, $B = 10$, $S = 8$ and $N = 11$ (the base). Thus the code $SEA$ is $871$.