In a pushdown automaton $P = (Q, \Sigma, \Gamma, \delta, q_0, F)$, a transition of the form, where $p, q \in Q$, $a \in \Sigma \cup \{\epsilon\}$, and $X, Y \in \Gamma \cup \{\epsilon\}$, represents $$(q, Y) \in \delta(p, a, X).$$ Consider the following pushdown automaton over the input alphabet $\Sigma = \{a, b\}$ and stack alphabet $\Gamma = \{\#, A\}$. The number of strings of length 100 accepted by the above pushdown automaton is _____.

Let us analyze the language accepted by the given pushdown automaton (PDA):

1. Initial transition: From the start state $q_0$, the transition $\epsilon, \epsilon \to \#$ moves to state $q_1$ and pushes the bottom-of-stack marker $\#$ onto the stack.

2. State $q_1$: The self-loop $a, \epsilon \to A$ reads input symbol $a$ and pushes $A$ onto the stack without popping anything. Thus, reading $n$ occurrences of $a$ pushes $n$ symbols $A$ onto the stack ($n \ge 0$).

3. Transition $q_1 \to q_2$: An $\epsilon$-transition $\epsilon, \epsilon \to \epsilon$ moves to state $q_2$ without altering the stack.

4. State $q_2$: The self-loop $b, A \to \epsilon$ reads input symbol $b$ and pops one symbol $A$ from the stack. If $m$ occurrences of $b$ are read, $m$ symbols $A$ are popped ($m \ge 0$).

5. Transition $q_2 \to q_3$ (Final state): The transition $\epsilon, A \to A$ moves to the accepting state $q_3$ without consuming any input, provided the top of the stack is $A$. This requires at least one $A$ to remain on top of the stack, which means:

   $$n - m \ge 1 \implies n > m$$

Thus, the PDA accepts strings of the form $w = a^n b^m$ such that $n > m \ge 0$.

We are asked to find the number of accepted strings of length $100$:

Substituting $n = 100 - m$ into the inequality $n > m$:

Since $m$ is an integer and $m \ge 0$, the possible values of $m$ are:

There are $50$ possible values for $m$, and each corresponds to a unique string $a^{100-m} b^m$.

Therefore, the number of accepted strings of length $100$ is 50.