jmc

We begin by replacing the coefficients and constants in the equation with their residues modulo 23. We find that 3874 divided by 23 gives a remainder of 10, 481 divided by 23 gives a remainder of 21, and 1205 gives a remainder of 9. So the given congruence is equivalent to $$10x+21\equiv 9(\mathrm{mod}23).$$Now add 2 to both sides to obtain $$10x\equiv 11(\mathrm{mod}23).$$Notice that we have replaced 23 with 0 on the left-hand side, since $23\equiv 0(\mathrm{mod}23)$. Now let us find the modular inverse of 10. We want to find an integer which is divisible by 10 and one more than a multiple of 23. Note that since the units digit of 23 is 3, the units digit of $3\times 23$ is 9, so $3\times 23+1$ is a multiple of 10. Thus $(3\times 23+1)/10=7$ is the modular inverse of 10. Multiplying both sides of $10x\equiv 11(\mathrm{mod}23)$ by 7 gives $x\equiv 77(\mathrm{mod}23)$, which implies $x\equiv 8(\mathrm{mod}23)$. So the three digit solutions are $$\begin{array}{rl}8+23\times 4& =100\\ 8+23\times 5& =123\\ & ⋮\\ 8+23\times 43& =997,\end{array}$$of which there are $\boxed{40}$.