The answer to your question has a very easy part and a hard part. First, the very easy part. You pose
$$ a x + b \cong n \pmod{100} $$
with the example
$$ 65 x + 6 \cong 81 \pmod{100} \text{.} $$
The basic strategy is the same one you would use if not reduction modulo $100$ were intended: use algebra to solve for $x$. The first step is very easy, subtract the constant from both sides.
\begin{align*}
a x &\cong n - b\pmod{100} \\
65 x &\cong 81 - 6 \\
&\cong 75 \pmod{100} \text{.}
\end{align*}
(I suspect that you will like to remember, if the result of this subtraction is negative, to add the modulus to the difference to get a result in the interval $[0,100)$ again.)
Now our strategy calls for division by $65$, but there is a "simpler" problem hiding in your example : $65$, $75$, and $100$ have a common factor, $5$, and we can reduce by this common factor.
$$ 13 x \cong 15 \pmod{20} \text{.} $$
More generally, if $a$ and the modulus, $100$ have a common factor then either $n-b$ also shares that factor and there is a solution or $n-b$ does not share that factor and there is no solution. So we check that any common factor of $a$ and $100$, $d$, is also a factor of $n-b$. Having factored $d$ out, we have
$$ (a/d)x \cong (n-b)/d \pmod{100/d} \text{.} $$
The easiest way to compute $d$ is using the Euclidean algorithm to compute the greatest common divisor (gcd) of $a$ and $100$. Having found and divided out the greatest common divisor, we know that $a/d$ and $100/d$ are relatively prime. This ensures that there is a number, $c$, such that
$$ c \cdot (a/d) \cong 1 \pmod{100/d} \text{,} $$
That is, $c$ acts like the reciprocal of the coefficient of $x$, so multplying by $c$ is like dividing by that coefficient.
How do we find this $c$? Use the extended Euclidean algorithm. The input is two integers, $a/d$ and $100/d$, whose gcd you want to know. The output is that gcd and also a pair of numbers, $e$ and $f$, with the following immediately useful property:
$$ e \cdot (a/d) + f \cdot (100/d) = \gcd(a/d,100/d) = 1 \text{.} $$
That is $e$ and $f$ are coefficients for each input that act like their reciprocals when using the other input as modulus. For us, we want $e$ and we multiply both sides of the equation by $e$:
\begin{align*}
\underbrace{e \cdot (a/d)}_{\cong 1 \pmod{100/d}} x &\cong e \cdot (n-b)/d \pmod{100/d} \\
x &\cong e \cdot (n-b)/d \pmod{100/d} \text{.}
\end{align*}
What does the computation of $e$look like in the example? We want to study the gcd of $13$ and $20$.
\begin{align*}
20 &= 1 \cdot 13 + 7 & 7 &= 1 \cdot 20 - 1 \cdot 13 \\
13 &= 1 \cdot 7 + 6 & 6 &= -1 \cdot 20 + 2 \cdot 13 \\
7 &= 1 \cdot 6 + 1 & 1 &= (7-6) = 2 \cdot 20 -3 \cdot 13 \text{.}
\end{align*}
(In the last line, we have reminded that the multiples of $20$ and $13$ we need are present on prior lines.)
This says that $2$ is the reciprocal of $20$ modulo $13$ and also that $-3$ is the reciprocal of $13$ modulo $20$. We want the latter and we add the modulus, $20$, to it to get a small positive number.
$$ 17 \cdot 13 \cong 1 \pmod{20} $$
So,
\begin{align*}
17 \cdot 13 x &\cong 17 \cdot 15 \pmod{20} \\
x &\cong 17 \cdot 15 \pmod{20} \\
x &\cong 15 \pmod{20} \text{.}
\end{align*}
If it had been the case that $d = 1$, so we were not working with a reduced modulus, we would be done. However, when $d > 1$ there are multiple solutions. We know that $15$ is one solution and we know that $15$ plus multiples of $20$ are solutions. Since the original congruence is modulo $100$, we want to know all the "$15$ plus multiples of $20$" in $[0,100)$. (By division, there are $100/20 = 5$, and in general, there are $d$ of them, but let's write them down.)
They are $15$, $35$, $55$, $75$, and $95$. (If we add $20$ again, we "wrap around" back to $15$.)
Let's test these:
\begin{align*}
65(15) + 6 &\cong 975 + 6 \\
&\cong 981 \\
&\cong 81 \pmod{100} \text{,} \\
65(35) + 6 &\cong 2275 + 6 \\
&\cong 2281 \\
&\cong 81 \pmod{100} \text{,} \\
65(55) + 6 &\cong 3575 + 6 \\
&\cong 3581 \\
&\cong 81 \pmod{100} \text{,} \\
65(75) + 6 &\cong 4875 + 6 \\
&\cong 4881 \\
&\cong 81 \pmod{100} \text{, and} \\
65(95) + 6 &\cong 6175 + 6 \\
&\cong 6181 \\
&\cong 81 \pmod{100} \text{.}
\end{align*}
This shows that all the numbers we found are solutions. The argument laid out above shows that these are all of them.
