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Human: Let $n$ be a positive integer. If $d \equiv (3^{2n} + 4)^{-1} \pmod{9}$, what is the remainder when $d$ is divided by $9$?

Assistant: To find $d \equiv (3^{2n} + 4)^{-1} \pmod{9}$, we need to first simplify $3^{2n} + 4$ modulo $9$ and then find the multiplicative inverse of this expression modulo $9$.

**Step 1: Simplifying $3^{2n} + 4 \pmod{9}$**

We use the fact that $3^2 \equiv 1 \pmod{9}$, since $3^2 = 9 \equiv 0 \pmod{9}$ and thus $3^2 + 9k \equiv 1 \pmod{9}$ for any integer $k$. Raising both sides to any power $n$, we get $3^{2n} \equiv 1^n \equiv 1 \pmod{9}$.

So, we have $3^{2n} + 4 \equiv 1 + 4 \equiv 5 \pmod{9}$.

**Step 2: Finding the multiplicative inverse modulo $9$**

We want to find an integer $x$ such that $(3^{2n} + 4)x \equiv 1 \pmod{9}$. Based on step 1, we need to find an $x$ such that $5x \equiv 1 \pmod{9}$.

We can check multiples of $5$ until we find one that, when divided by $9$, leaves a remainder of $1$. We can quickly test these:

- $5 \cdot 1 \equiv 5 \pmod{9}$
- $5 \cdot 2 \equiv 10 \equiv 1 \pmod{9}$

So, $x = 2$ is such that $5x \equiv 1 \pmod{9}$, and thus $x = 2$ is the multiplicative inverse of $5$ modulo $9$.

Therefore, $d \equiv (3^{2n} + 4)^{-1} \equiv 2 \pmod{9}$.

The remainder when $d$ is divided by $9$ is $\boxed