help pls

To solve this congruence, we need to find the smallest positive integer $n$ such that $5^n \equiv n^5 \pmod{3}$.

First, let's observe that $5 \equiv 2 \pmod{3}$. Therefore, $5^n \equiv 2^n \pmod{3}$.

Now, let's calculate the values of $2^n$ modulo 3 for small values of $n$:

- $2^1 \equiv 2 \pmod{3}$

- $2^2 \equiv 1 \pmod{3}$

- $2^3 \equiv 2 \pmod{3}$

- $2^4 \equiv 1 \pmod{3}$

From this, we can see a pattern emerge: the value of $2^n$ alternates between 1 and 2 modulo 3, with period 2.

Now, let's consider $n^5$. Since we're taking $n^5$ modulo 3, we can reduce $n^5$ to its residue modulo 3:

- $1^5 \equiv 1 \pmod{3}$

- $2^5 \equiv 32 \equiv 2 \pmod{3}$

- $3^5 \equiv 243 \equiv 0 \pmod{3}$

The pattern for $n^5$ modulo 3 is not as obvious as $2^n$, but we can see that for $n = 2$, $n^5$ matches $2^n$ modulo 3.

So, the smallest positive integer $n$ such that $5^n \equiv n^5 \pmod{3}$ is $n = 2$.