Number System

Let $f(N) = (3N^2 + 8N - 3)(pN - 1)$. We need to find the remainder when $f(N)$ is divided by $N + 1$. According to the Remainder Theorem, the remainder of $f(N)$ when divided by $N + 1$ is equal to $f(-1)$.

To find $f(-1)$, we start by calculating $3(-1)^2 + 8(-1) - 3$:

$$3(-1)^2 + 8(-1) - 3 = 3(1) - 8 - 3 = 3 - 8 - 3 = -8$$

Next, we calculate $p(-1) - 1$:

$$p(-1) - 1 = -p - 1$$

Now we can find $f(-1)$:

$$f(-1) = (-8)(-p - 1) = 8(p + 1)$$

According to the problem, the remainder $f(-1)$ is given to be 24:

$$8(p + 1) = 24$$

Dividing both sides by 8, we find:

$$p + 1 = 3 \implies p = 2$$

Now we want to find the remainder when $p^4$ is divided by 10. Calculating $p^4$:

$$p^4 = 2^4 = 16$$

Now we find the remainder of 16 when divided by 10:

$$16 \mod 10 = 6$$

Thus, the remainder when $p^4$ is divided by 10 is:

$$\boxed{6}$$