Sherlock Holmes Base Number Problem 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 &quot;AT SEA BASE. SEAS EBB SEA: BASS. &quot;
Dr. Watson comments, &quot;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.&quot;
Holmes calmly turns the combination lock and opens the suitcase. While Dr. Watson gapes in surprise, Holmes replies, &quot;You were indeed right, and the answer was just the value of the word $SEA$, interpreted as decimal digits.&quot; 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!
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+&amp;Ex^2+&amp;Ax+&amp;S&amp;\\
     &amp;Ex^2+&amp;Bx+&amp;B&amp;\\
     &amp;Sx^2+&amp;Ex+&amp;A&amp;\quad+\\
\hline
Bx^3+&amp;Ax^2+&amp;Sx+&amp;S&amp;
\end{eqnarray*}
In each column we have at most $2$ carries, so from this it follows that
\begin{eqnarray*}
S+B+A &amp;=&amp; S+k_1x,\\
A+B+E+k_1&amp;=&amp; S+k_2x,\\
E+E+S+k_2&amp;=&amp; A+k_3x,\\
S+k_3 &amp;=&amp; B,
\end{eqnarray*}
where $0\leq k_1,k_2,k_3\leq2$, and $0\leq A,B,E,S&lt;x$ are distinct.
The first equation shows that $B+A=k_1x$, so $k_1=1$ because $0\leq A,B&lt;x$ 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 $0\leq E&lt;x$ 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&amp;=&amp;\frac{3x+8}{4}&amp;=&amp;\frac34x+2,\\
A&amp;=&amp;x-B&amp;=&amp;\frac14x-2,\\
S&amp;=&amp;B-2&amp;=&amp;\frac34x,\\
E&amp;=&amp;B-3&amp;=&amp;\frac34x-1,\\
\end{eqnarray*}
which shows that $x$ is divisible by $4$ and $x\geq12$ because $B&lt;x$. Because $S,E,A&lt;10$ it follows that $x=12$, and $(S,E,A)=(8,7,1)$.
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}
&amp;S&amp;E&amp;A&amp;S\\
+&amp;&amp;E&amp;B&amp;B\\
+&amp;&amp;S&amp;E&amp;A\\
\hline
&amp;B&amp;A&amp;S&amp;S
\end{matrix}
which gives successively

$S + B + A = S + N$, whence $A + B = N$ and $A &gt; 0$,
$1 + A + B + E = S + N$, whence $S = E + 1$
(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 $A &gt; 0$, one has necessarily $A = 1$, $E = 7$, $B = 10$, $S = 8$ and $N = 11$ (the base). Thus the code $SEA$ is $871$.