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By the prime number theorem, the nth prime is like n*log n.  This is not a polynomial, so there must be an n such that p(n) is composite.

The answer is 4810.

The answer is 25.

The answer is 14.

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.