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By the Binomial Theorem, the coefficient is 4500.

There are 540 ways.

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:

\[
\boxed{6}
\]

We can simplify the given equation: 5a + 3b + 5c = 4a + 5b + 4c a - 2b - c = 0 a = 2b + c

Now we can substitute a in the expression for N: N = 5(2b + c) + 3b + 5c N = 13b + 10c

We are given that 131 < N < 150. Let's try some values for b and c:

If b = 10 and c = 1, then N = 140, which is within the given range.

If b = 11 and c = 1, then N = 153, which is outside the given range.

Therefore, the maximum possible value of b is 10.

Now we can find the maximum possible value of a + 2b + 3c: a + 2b + 3c = (2b + c) + 2b + 3c = 5b + 4c

Substituting b = 10 and c = 1, we get: 5b + 4c = 5(10) + 4(1) = 54

Therefore, the maximum possible value of a + 2b + 3c is 54.

The six teams play C(6,2) = 15 games.

Therefore, the probability that at least two different teams finish the tournament with the same number of wins is 1 - 15/2^15 = 32753/32768.

The number of solutions is 15.

The numbers that work are 0, 1, 40, 60, 80, 100, 120, 140, and 160.

Using inverses, we get 5n - 3.