$33 \cdot 70$ A cryptographer designed the following method for encoding natural numbers: First, represent the natural number in base 5, then establish a one-to-one correspondence between the digits in the base 5 representation and the elements of the set $\{V, W, X, Y, Z\}$. Using this correspondence, he found that three consecutive increasing natural numbers were encoded as $V Y Z, V Y X, V V W$. The natural number encoded as $X Y Z$ in decimal notation is
(A) 48.
(B) 71.
(C) 82.
(D) 108.
(E) 113.
(38th American High School Mathematics Examination, 1987)
[Solution] Given that $V Y Z, V Y X, V V W$ are three consecutive increasing numbers, then
$$
\left\{\begin{array}{l}
V Y Z+1=V Y X, \\
V Y X+1=V V W .
\end{array}\right.
$$

Obviously, $Z \neq 1,2$, otherwise from (1) we have $X=2,3$,
which makes (2) not hold, because at this time $X+1=3$ or 4, and no carry-over problem would occur.
Thus, $Z=3, X=4, W=0$.
Also, $Y \neq 2$, otherwise $V Y X+1=V 24+1=V 30=V 2 W$, which contradicts the given condition! Therefore, $Y=1$.
$\therefore X=4, Y=1, Z=3$. That is, $X Y Z=(413)_{5}$.
And $\quad(413)_{5}=4 \cdot 5^{2}+1 \cdot 5^{1}+3=108$.
Therefore, the answer is $(D)$.