33.61 When a number $N$ is expressed in base 7, it is a three-digit number, and when it is expressed in base 9, its three digits are exactly reversed. Then the middle digit of $N$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) 5.
(19th American High School Mathematics Examination, 1968)
[Solution] Let the three digits of $N$ in base 9 be $x, y, z$, i.e., $9^{2} x+9^{1} y + z$. It is also given that its base 7 representation is $7^{2} z+7^{1} y+x$.
Thus, $81 x+9 y+z=49 z+7 y+x$, or $y=8(3 z-5 x)$.
Since $y$ is a digit in the base 7 representation of $N$, then $0 \leqslant y < 7$.
Therefore, the middle digit of $N$ is 0.
Since $z$ is also the leading digit in the base 7 representation of $N$, we have $0<z<7$.

And $3 z=5 x$, so $z$ must be divisible by 5, hence $z=5$, and simultaneously $x=3$.
Thus, $N=305_{9}=503_{7}=248_{10}$.
Therefore, the answer is $(A)$.