$1 \cdot 30$ Find the largest natural number $x$, such that for every natural number $y, x$ divides
$$
7^{y}+12 y-1 \text {. }
$$
("Friendship Cup" International Mathematics Competition, 1992)
[Solution] When $y=1$, $7^{y}+12 y-1=18$, therefore
$$
\begin{array}{l}
x \mid 18 . \\
x \leqslant 18 .
\end{array}
$$

Thus
we will use mathematical induction to prove
$$
18 \mid 7^{y}+12 y-1 .
$$

When $y=1$, it is obvious.
If $y=k$, then $18 \mid 7^{k}+12 k-1$.
If $y=k+1$, then
$$
\begin{aligned}
& 7^{k+1}+12(k+1)-1 \\
= & 7 \cdot 7^{k}+7 \cdot 12 k-7-6 \cdot 12 k+18 \\
= & 7\left(7^{k}+12 k-1\right)-72 k+18 .
\end{aligned}
$$

By the induction hypothesis

Also
$$
\begin{array}{c}
18 \mid 7^{k}+12 k-1, \\
18 \mid-72 k+18
\end{array}
$$

Therefore
$$
18 \mid 7^{k+1}+12(k+1)-1 .
$$

That is, the proposition holds for $y=k+1$.
Thus, for all natural numbers $y, \quad 18 \mid 7^{y}+12 y-1$.
Hence, the largest natural number $x=18$.