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:

\[
\boxed{6}
\]