jmc

We can start by multiplying both sides of the congruence by $10$ and evaluating both sides modulo $997$: $$\begin{array}{rl}10\cdot 100x& \equiv 10\cdot 1(\mathrm{mod}997)\\ 1000x& \equiv 10(\mathrm{mod}997)\\ 3x& \equiv 10(\mathrm{mod}997)\end{array}$$

Why multiply by $10$? Well, as the computations above show, the result is to produce a congruence equivalent to the original congruence, but with a much smaller coefficient for $x$.

From here, we could repeat the same strategy a couple more times; for example, multiplying both sides by $333$ would give $999x\equiv 2x$ on the left side, reducing the coefficient of $x$ further. One more such step would reduce the coefficient of $x$ to $1$, giving us the solution.

However, there is an alternative way of solving $3x\equiv 10(\mathrm{mod}997)$. We note that we can rewrite this congruence as $3x\equiv -987(\mathrm{mod}997)$ (since $10\equiv -987(\mathrm{mod}997)$). Then $-987$ is a multiple of $3$: specifically, $-987=3\cdot (-329)$, so multiplying both sides by $3^{-1}$ gives $$x\equiv -329(\mathrm{mod}997).$$ This is the solution set to the original congruence. The unique three-digit positive solution is $$x=-329+997=\boxed{668}.$$