33.68 If the octal representation of a perfect square is $a b 3 c$ (i.e., $n^{2}=$ $\left.(a b 3 c)_{8}=8^{3} a+8^{2} b+8 \cdot 3+c\right)$, where $a \neq 0$, then $c$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) cannot be uniquely determined.
(33rd American High School Mathematics Examination, 1982)
[Solution 1] If $n^{2}=(a b 3 c)_{8}$, let $n=(d e)_{8}$, then
$$
n^{2}=(8 d+e)^{2}=64 d^{2}+8(2 d e)+e^{2} \text {. }
$$

Therefore, 3 is the sum of the second digit (counting from the right, same below) of $e^{2}$ in octal and the first digit of $2 d e$ (the sum in octal sense). The latter is an even number, and 3 is odd, so the second digit of $e^{2}$ in octal must be odd.
Below we list the square table in octal:
\begin{tabular}{c|c|c|c|c|c|c|c}
\hline$e$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline$e^{2}$ & 1 & 4 & 11 & 20 & 31 & 44 & 61 \\
\hline
\end{tabular}

Among these, the second digit of $e^{2}$ is odd only when $e$ is 3 or 5. In either case, the first digit of $e^{2}$ in octal is 1, i.e., $c=1$.
Therefore, the answer is $(B)$.
[Solution 2] According to the problem, $n^{2}=(a b 3 c)_{8}=8^{3} a+8^{2} b+8 \cdot 3+c$ (note that $n^{2}$ here is a decimal number), then the remainder of $n^{2}$ divided by 8 must be 0, 1, or 4, because
if $n=4 k$, then $n^{2}=16 k^{2}$ is divisible by 8 with a remainder of 0;
if $n=4 k+1$, then $n^{2}=16 k^{2}+8 k+1$, which is divisible by 8 with a remainder of 1;
if $n=4 k+2$, then $n^{2}=16 k^{2}+16 k+4$, which is divisible by 8 with a remainder of 4;
if $n=4 k+3$, then $n^{2}=16 k^{2}+24 k+9$, which is divisible by 8 with a remainder of 1.
Thus, $n^{2} \equiv 0,1$ or $4(\bmod 8)$.
Therefore, $c=0$, or $c=1$, or $c=4$.
If $c=0$, then $n^{2}=8(8 k+3)=4(16 k+6)$. Since 4 is a perfect square, $16 k+6$ must also be a perfect square, but $16 k+6$ leaves a remainder of 6 when divided by 8, which is impossible, so $c \neq 0$.

If $c=4$, then $n^{2}=4(8 l+7)$, similarly, $8 l+7$ leaves a remainder of 7 when divided by 8, so $8 l+7$ cannot be a perfect square, thus $c \neq 4$.
Therefore, $c$ can only be 1.

Hence, the answer is $(B)$.