[User] Given the equation \(\frac{31}{195} = \frac{1}{A + \frac{1}{B + \frac{1}{C + \frac{1}{D}}}}\), where \(A, B, C, D\) are positive integers, find \(A^2 + B^2 + C^2 + D^2\).

[Assistant]
To solve the equation \(\frac{31}{195} = \frac{1}{A + \frac{1}{B + \frac{1}{C + \frac{1}{D}}}}\), where \(A, B, C, D\) are positive integers, we need to express \(\frac{31}{195}\) as a continued fraction.

First, take the reciprocal of \(\frac{31}{195}\):
\[
\frac{195}{31} = 6 + \frac{9}{31}
\]
This gives \(A = 6\), and we need to find the continued fraction for \(\frac{9}{31}\).

Next, take the reciprocal of \(\frac{9}{31}\):
\[
\frac{31}{9} = 3 + \frac{4}{9}
\]
This gives \(B = 3\), and we need to find the continued fraction for \(\frac{4}{9}\).

Next, take the reciprocal of \(\frac{4}{9}\):
\[
\frac{9}{4} = 2 + \frac{1}{4}
\]
This gives \(C = 2\), and we need to find the continued fraction for \(\frac{1}{4}\).

Finally, take the reciprocal of \(\frac{1}{4}\):
\[
\frac{4}{1} = 4
\]
This gives \(D = 4\).

Thus, the continued fraction representation is:
\[
\frac{31}{195} = \frac{1}{6 + \frac{1}{3 + \frac{1}{2 + \frac{1}{4}}}}
\]

Now, we need to find \(A^2 + B^2 + C^2 + D^2\):
\[
A = 6, \quad B = 3, \quad C = 2, \quad D = 4
\]
\[
A^2 + B^2 + C^2 + D^2 = 6^2 + 3^2 + 2^2 + 4^2 = 36 + 9 + 4 + 16 = 65
\]

Therefore, the final answer is:
\[
\boxed{65}
\]

System: You are a helpful assistant.


[User] Let \( d_k(n) \) denote the number of ways to write \( n \) as a product of \( k \) factors. Let \( f(x) = \sum_{n=1}^{\infty} \frac{d_x(n)}{n^{2022}} \). Find \( \log_{f(16)} f(64) \).

[Assistant]
To solve the problem, we need to find \(\log_{f(16)} f(64)\), where \(f(x) = \sum_{n=1}^{\infty} \frac{d_x(n)}{n^{2022}}\).

First, recall that \(d_k(n)\) is the number of ways to write \(n\) as a product of \(k\) factors. The Dirichlet generating function for \(d_k(n)\) is \(\zeta(s)^k\), where \(\zeta(s)\) is the Riemann zeta function. This is because each prime factor's contribution to the generating function is \((1 - p^{-s})^{-k}\), and the product over all primes gives \(\zeta(s)^k\).

Given \(f(x) = \sum_{n=1}^{\infty} \frac{d_x(n)}{n^{2022}}\), we recognize this as the Dirichlet generating function evaluated at \(s = 2022\). Therefore, we have:
\[
f(x) = \zeta(2022)^x
\]

Next, we need to compute \(\log_{f(16)} f(64)\). Substituting the expressions for \(f(16)\) and \(f(64)\), we get:
\[
f(16) = \zeta(2022)^{16} \quad \text{and} \quad f(64) = \zeta(2022)^{64}
\]

Using the properties of logarithms, we find:
\[
\log_{\zeta(2022)^{16}} (\zeta(2022)^{64}) = \frac{\ln (\zeta(2022)^{64})}{\ln (\zeta(2022)^{16})} = \frac{64 \ln \zeta(2022)}{16 \ln \zeta(2022)} = \frac{64}{16} = 4
\]

Thus, the final answer is:
\[
\boxed{4}
\]

System: You are a helpful assistant.


[User] The probability that Kim has a math test today is $\frac{4}{7}$. What is the probability that Kim does not have a math test today? Express your answer as a common fraction.

[Assistant]
To find the probability that Kim does not have a math test today, we start with the given probability that Kim has a math test today, which is \(\frac{4}{7}\).

The probability of the complementary event (not having a math test) is calculated by subtracting the probability of having a math test from 1. This is because the total probability of all possible outcomes must sum to 1.

\[
\text{Probability of not having a math test} = 1 - \text{Probability of having a math test}
\]

Substituting the given probability:

\[
\text{Probability of not having a math test} = 1 - \frac{4}{7}
\]

To perform the subtraction, we express 1 as \(\frac{7}{7}\):

\[
1 - \frac{4}{7} = \frac{7}{7} - \frac{4}{7} = \frac{3}{7}
\]

Thus, the probability that Kim does not have a math test today is:

\[
\boxed{\frac{3}{7}}
\]

System: You are a helpful assistant.


[User] If $x$ is a positive integer, what is the value of $x$ for the equation $(x!-(x-3)!) \div 23 = 1$?

[Assistant]
To solve the equation \((x! - (x-3)!) \div 23 = 1\) for a positive integer \(x\), we start by rewriting the equation to eliminate the denominator:

\[
x! - (x-3)! = 23
\]

Next, we need to find a positive integer \(x\) that satisfies this equation. We will test small values of \(x\) to see if they work.

1. **Testing \(x = 3\):**
   \[
   (x-3)! = 0! = 1
   \]
   \[
   x! = 3! = 6
   \]
   \[
   x! - (x-3)! = 6 - 1 = 5
   \]
   Since \(5 \neq 23\), \(x = 3\) is not a solution.

2. **Testing \(x = 4\):**
   \[
   (x-3)! = 1! = 1
   \]
   \[
   x! = 4! = 24
   \]
   \[
   x! - (x-3)! = 24 - 1 = 23
   \]
   Since \(23 = 23\), \(x = 4\) is a solution.

3. **Testing \(x = 5\):**
   \[
   (x-3)! = 2! = 2
   \]
   \[
   x! = 5! = 120
   \]
   \[
   x! - (x-3)! = 120 - 2 = 118
   \]
   Since \(118 \neq 23\), \(x = 5\) is not a solution.

4. **Testing \(x = 6\):**
   \[
   (x-3)! = 3! = 6
   \]
   \[
   x! = 6! = 720
   \]
   \[
   x! - (x-3)! = 720 - 6 = 714
   \]
   Since \(714 \neq 23\), \(x = 6\) is not a solution.

From the above calculations, we see that the only positive integer \(x\) that satisfies the equation \(x! - (x-3)! = 23\) is \(x = 4\).

Thus, the solution is:
\[
\boxed{4}
\]

System: You are a helpful assistant.


[User] Find the number of quadruples $(x, y, z, w)$ of non-negative integers that satisfy the inequality $w + x + y + z \le 1992$.

[Assistant]
To find the number of quadruples \((w, x, y, z)\) of non-negative integers that satisfy the inequality \(w + x + y + z \leq 1992\), we can introduce a slack variable \(t\) such that \(w + x + y + z + t = 1992\). This converts the inequality into an equality, where \(t\) is a non-negative integer representing the "leftover" from the maximum sum of 1992.

Now, we need to find the number of non-negative integer solutions to the equation \(w + x + y + z + t = 1992\). This is a classic application of the stars and bars method. The number of solutions to the equation \(a_1 + a_2 + \cdots + a_k = n\) in non-negative integers is given by the binomial coefficient \(\binom{n + k - 1}{k - 1}\).

In our case, \(n = 1992\) and \(k = 5\) (since we have 5 variables \(w, x, y, z, t\)). Therefore, the number of solutions is:

\[
\binom{1992 + 5 - 1}{5 - 1} = \binom{1996}{4}
\]

Thus, the number of quadruples \((w, x, y, z)\) of non-negative integers that satisfy the inequality \(w + x + y + z \leq 1992\) is \(\boxed{\binom{1996}{4}}\).

System: You are a helpful assistant.


[User] Evaluate the integral \(\int \log(x + \sqrt{x^2 + a^2}) \, dx\).

[Assistant]
To evaluate the integral \(\int \log(x + \sqrt{x^2 + a^2}) \, dx\), we use integration by parts. Let:

\[
u = \log(x + \sqrt{x^2 + a^2}) \quad \text{and} \quad dv = dx
\]

Then, we need to find \(du\) and \(v\):

\[
du = \frac{d}{dx} \left( \log(x + \sqrt{x^2 + a^2}) \right) dx
\]

To find \(du\), we use the chain rule:

\[
\frac{d}{dx} \left( \log(x + \sqrt{x^2 + a^2}) \right) = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right)
\]

The derivative of \(x + \sqrt{x^2 + a^2}\) is:

\[
\frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right) = 1 + \frac{x}{\sqrt{x^2 + a^2}}
\]

Thus,

\[
du = \frac{1 + \frac{x}{\sqrt{x^2 + a^2}}}{x + \sqrt{x^2 + a^2}} dx = \frac{\sqrt{x^2 + a^2} + x}{(x + \sqrt{x^2 + a^2})\sqrt{x^2 + a^2}} dx = \frac{1}{\sqrt{x^2 + a^2}} dx
\]

And,

\[
v = \int dx = x
\]

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), we get:

\[
\int \log(x + \sqrt{x^2 + a^2}) \, dx = x \log(x + \sqrt{x^2 + a^2}) - \int x \cdot \frac{1}{\sqrt{x^2 + a^2}} \, dx
\]

Next, we evaluate the remaining integral \(\int \frac{x}{\sqrt{x^2 + a^2}} \, dx\). We use the substitution \(w = x^2 + a^2\), so \(dw = 2x \, dx\) or \(x \, dx = \frac{1}{2} dw\):

\[
\int \frac{x}{\sqrt{x^2 + a^2}} \, dx = \int \frac{1}{2} \frac{dw}{\sqrt{w}} = \frac{1}{2} \int w^{-\frac{1}{2}} \, dw = \frac{1}{2} \cdot 2 \sqrt{w} + C = \sqrt{w} + C = \sqrt{x^2 + a^2} + C
\]

Combining the results, we have:

\[
\int \log(x + \sqrt{x^2 + a^2}) \, dx = x \log(x + \sqrt{x^2 + a^2}) - \sqrt{x^2 + a^2} + C
\]

Thus, the final answer is:

\[
\boxed{x \log\left(x + \sqrt{x^2 + a^2}\right) - \sqrt{x^2 + a^2} + C}
\]

System: You are a helpful assistant.


[User] A recent study found that 60$\%$ of men and 80$\%$ of women surveyed support increased funding for particular medical research. The study surveyed 100 men and 900 women. What was the overall percent of the people surveyed who supported increased funding?

[Assistant]
To find the overall percentage of people surveyed who supported increased funding for medical research, we need to follow these steps:

1. Calculate the number of men who supported the funding.
2. Calculate the number of women who supported the funding.
3. Sum the supporters from both groups to get the total number of supporters.
4. Calculate the overall percentage of supporters.

Given:
- 60% of 100 men supported the funding.
- 80% of 900 women supported the funding.
- Total number of people surveyed = 100 men + 900 women = 1000 people.

Step 1: Calculate the number of men who supported the funding.
\[ \text{Number of men supporting} = 60\% \times 100 = 0.60 \times 100 = 60 \]

Step 2: Calculate the number of women who supported the funding.
\[ \text{Number of women supporting} = 80\% \times 900 = 0.80 \times 900 = 720 \]

Step 3: Sum the supporters from both groups.
\[ \text{Total number of supporters} = 60 + 720 = 780 \]

Step 4: Calculate the overall percentage of supporters.
\[ \text{Overall percentage} = \left( \frac{780}{1000} \right) \times 100 = 78\% \]

Thus, the overall percentage of people surveyed who supported increased funding is \(\boxed{78\%}\).

System: You are a helpful assistant.


[User] For some positive integer $k$, when 60 is divided by $k^2$, the remainder is 6. What is the remainder when 100 is divided by $k$?

[Assistant]
To solve the problem, we start by noting that when 60 is divided by \( k^2 \), the remainder is 6. This can be expressed mathematically as:
\[ 60 = k^2 \cdot q + 6 \]
for some integer \( q \). Rearranging this equation, we get:
\[ 60 - 6 = k^2 \cdot q \]
\[ 54 = k^2 \cdot q \]
This implies that \( k^2 \) must be a divisor of 54. The divisors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. Since \( k \) is a positive integer, \( k^2 \) must also be a positive integer and a perfect square. The perfect squares among the divisors of 54 are 1 and 9. However, since the remainder (6) must be less than the divisor \( k^2 \), \( k^2 \) must be greater than 6. Therefore, the only possible value for \( k^2 \) is 9, which means \( k = 3 \).

Next, we need to find the remainder when 100 is divided by \( k \). Substituting \( k = 3 \), we calculate:
\[ 100 \div 3 = 33 \text{ remainder } 1 \]
This can be verified by the calculation:
\[ 100 = 3 \cdot 33 + 1 \]
Thus, the remainder when 100 is divided by 3 is 1.

Therefore, the answer is:
\[
\boxed{1}
\]

System: You are a helpful assistant.


[User] Find the surface area of a sphere with center $(5,3,4)$ that is tangent to the plane $2x+2y-z=-6$.

[Assistant]
To find the surface area of a sphere with center \((5,3,4)\) that is tangent to the plane \(2x + 2y - z = -6\), we first need to determine the radius of the sphere. The radius is the distance from the center of the sphere to the plane.

The formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(ax + by + cz + d = 0\) is given by:
\[
D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}
\]

For the given plane \(2x + 2y - z = -6\), we can rewrite it in the standard form \(2x + 2y - z + 6 = 0\). Here, \(a = 2\), \(b = 2\), \(c = -1\), and \(d = 6\).

The center of the sphere is \((5, 3, 4)\). Plugging these values into the distance formula, we get:
\[
D = \frac{|2(5) + 2(3) - 1(4) + 6|}{\sqrt{2^2 + 2^2 + (-1)^2}}
\]

Calculating the numerator:
\[
2(5) + 2(3) - 1(4) + 6 = 10 + 6 - 4 + 6 = 18
\]

Calculating the denominator:
\[
\sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3
\]

Thus, the distance \(D\) is:
\[
D = \frac{18}{3} = 6
\]

The radius \(r\) of the sphere is 6. The surface area \(A\) of a sphere is given by:
\[
A = 4\pi r^2
\]

Substituting \(r = 6\):
\[
A = 4\pi (6)^2 = 4\pi \cdot 36 = 144\pi
\]

Therefore, the surface area of the sphere is:
\[
\boxed{144\pi}
\]

System: You are a helpful assistant.


[User] Compute $\sum_{r=1}^{n} r(r-1) \binom{n}{r}$.

[Assistant]
To compute the sum \(\sum_{r=1}^{n} r(r-1) \binom{n}{r}\), we start by expressing \(r(r-1) \binom{n}{r}\) in a more convenient form.

First, recall the binomial coefficient \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). We can rewrite \(r(r-1) \binom{n}{r}\) as follows:
\[
r(r-1) \binom{n}{r} = r(r-1) \cdot \frac{n!}{r!(n-r)!}.
\]
Notice that \(r(r-1) = r(r-1) \cdot 1 = r(r-1) \cdot \frac{(r-2)!}{(r-2)!} = \frac{r(r-1)(r-2)!}{(r-2)!} = \frac{r!}{(r-2)!}\). Thus,
\[
r(r-1) \binom{n}{r} = \frac{r!}{(r-2)!} \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-2)!(n-r)!} = n(n-1) \cdot \frac{(n-2)!}{(r-2)!(n-r)!} = n(n-1) \binom{n-2}{r-2}.
\]

Therefore, the original sum can be rewritten as:
\[
\sum_{r=1}^{n} r(r-1) \binom{n}{r} = \sum_{r=1}^{n} n(n-1) \binom{n-2}{r-2}.
\]
We can factor out the constant \(n(n-1)\) from the sum:
\[
n(n-1) \sum_{r=1}^{n} \binom{n-2}{r-2}.
\]
Next, we shift the index of summation by letting \(k = r-2\). When \(r = 1\), \(k = -1\) (but \(\binom{n-2}{-1} = 0\)), and when \(r = n\), \(k = n-2\). Therefore, the sum becomes:
\[
n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k}.
\]
The sum \(\sum_{k=0}^{n-2} \binom{n-2}{k}\) is the binomial expansion of \((1+1)^{n-2} = 2^{n-2}\). Thus, we have:
\[
n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k} = n(n-1) \cdot 2^{n-2}.
\]

Therefore, the value of the sum is:
\[
\boxed{n(n-1)2^{n-2}}.
\]

System: You are a helpful assistant.


[User] Let \( f(x) = \ln \left( \int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt \right) \). Evaluate \( f'(x) \) in terms of \( x \).

[Assistant]
To find \( f'(x) \) where \( f(x) = \ln \left( \int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt \right) \), we proceed as follows:

1. Define \( F(x) = \int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt \).
2. Then, \( f(x) = \ln(F(x)) \).

Using the chain rule, the derivative of \( f(x) \) is:
\[
f'(x) = \frac{F'(x)}{F(x)}.
\]

Next, we need to compute \( F'(x) \). By differentiating under the integral sign, we get:
\[
F'(x) = \frac{d}{dx} \left( \int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt \right) = \int_{0}^{\frac{\pi}{6}} \frac{\partial}{\partial x} \left( (\sin t)^x \right) \, dt.
\]

The partial derivative of \( (\sin t)^x \) with respect to \( x \) is:
\[
\frac{\partial}{\partial x} \left( (\sin t)^x \right) = (\sin t)^x \ln(\sin t).
\]

Therefore,
\[
F'(x) = \int_{0}^{\frac{\pi}{6}} (\sin t)^x \ln(\sin t) \, dt.
\]

Substituting \( F'(x) \) back into the expression for \( f'(x) \), we obtain:
\[
f'(x) = \frac{\int_{0}^{\frac{\pi}{6}} (\sin t)^x \ln(\sin t) \, dt}{\int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt}.
\]

Thus, the final answer is:
\[
\boxed{\frac{\int_{0}^{\frac{\pi}{6}} (\sin t)^x \ln(\sin t) \, dt}{\int_{0}^{\frac{\pi}{6}} (\sin t)^x \, dt}}.
\]

System: You are a helpful assistant.


[User] Given the general term of a sequence \( t_n = 3 \cdot 5^{n-1} \), derive the recursive formula for the sequence.

[Assistant]
Given the general term of the sequence \( t_n = 3 \cdot 5^{n-1} \), we need to derive the recursive formula for the sequence.

1. **Identify the type of sequence:**
   The given sequence \( t_n = 3 \cdot 5^{n-1} \) is a geometric sequence with the first term \( t_1 = 3 \) and common ratio \( r = 5 \).

2. **Determine the recursive formula:**
   In a geometric sequence, each term is obtained by multiplying the previous term by the common ratio. Therefore, the recursive formula is:
   \[
   t_n = r \cdot t_{n-1}
   \]
   Substituting the common ratio \( r = 5 \), we get:
   \[
   t_n = 5 \cdot t_{n-1}
   \]

3. **Specify the initial term:**
   The initial term of the sequence is:
   \[
   t_1 = 3
   \]

4. **Combine the initial term and the recursive relation:**
   The recursive formula for the sequence is:
   \[
   t_1 = 3
   \]
   \[
   t_n = 5 \cdot t_{n-1} \quad \text{for} \quad n \geq 2
   \]

Thus, the recursive formula for the sequence is:
\[
\boxed{t_1 = 3 \text{ and } t_n = 5t_{n-1} \text{ for } n \geq 2}
\]

System: You are a helpful assistant.


[User] If \(a\) is real, what is the only real number that could be a multiple root of the equation \(x^3 + ax + 1 = 0\)?

[Assistant]
To determine the only real number that could be a multiple root of the equation \(x^3 + ax + 1 = 0\), we use the fact that a multiple root must satisfy both the polynomial and its derivative.

Given the polynomial \(f(x) = x^3 + ax + 1\), its derivative is \(f'(x) = 3x^2 + a\). If \(r\) is a multiple root, it must satisfy the following system of equations:
1. \(r^3 + ar + 1 = 0\)
2. \(3r^2 + a = 0\)

From the second equation, solve for \(a\):
\[
a = -3r^2
\]

Substitute \(a\) into the first equation:
\[
r^3 + (-3r^2)r + 1 = 0
\]
\[
r^3 - 3r^3 + 1 = 0
\]
\[
-2r^3 + 1 = 0
\]
\[
r^3 = \frac{1}{2}
\]

The real solution to this equation is the cube root of \(\frac{1}{2}\):
\[
r = \sqrt[3]{\frac{1}{2}}
\]

Thus, the only real number that could be a multiple root is \(\boxed{\sqrt[3]{\frac{1}{2}}}\).

System: You are a helpful assistant.


[User] Find \(\lim_{x \to \infty} \frac{\sin(x)}{x}\).

[Assistant]
To find the limit \(\lim_{x \to \infty} \frac{\sin(x)}{x}\), we can use the Squeeze Theorem. The Squeeze Theorem states that if \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in some interval, and if \(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} h(x) = L\), then \(\lim_{x \to \infty} g(x) = L\).

1. **Bounding the Sine Function**: We know that \(-1 \leq \sin(x) \leq 1\) for all \(x\).

2. **Dividing by \(x\)**: Since \(x\) is positive as \(x \to \infty\), we can divide the inequality by \(x\):
   \[
   -\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}.
   \]

3. **Evaluating the Limits**: As \(x \to \infty\),
   \[
   \lim_{x \to \infty} -\frac{1}{x} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{1}{x} = 0.
   \]

4. **Applying the Squeeze Theorem**: Since both \(-\frac{1}{x}\) and \(\frac{1}{x}\) approach 0 as \(x \to \infty\), by the Squeeze Theorem, we have:
   \[
   \lim_{x \to \infty} \frac{\sin(x)}{x} = 0.
   \]

Therefore, the limit is:
\[
\boxed{0}.
\]

System: You are a helpful assistant.


[User] Determine if the function \( g: \mathbb{N} \to \mathbb{N} \) defined by \( g(x) = \left\lfloor \frac{3x+1}{3} \right\rfloor \) is bijective and find its inverse if it exists.

System: You are a helpful assistant.


[User] The median of the set $\{n, n + 5, n + 6, n + 9, n + 15\}$ is 9. What is the mean?

[Assistant]
To find the mean of the set \(\{n, n + 5, n + 6, n + 9, n + 15\}\) given that the median is 9, we proceed as follows:

1. **Identify the median**: The median of a set of 5 numbers is the third number when the set is ordered. For the set \(\{n, n + 5, n + 6, n + 9, n + 15\}\), the third term is \(n + 6\). Given that the median is 9, we have:
   \[
   n + 6 = 9
   \]
   Solving for \(n\):
   \[
   n = 9 - 6 = 3
   \]

2. **Substitute \(n\) back into the set**: With \(n = 3\), the set becomes:
   \[
   \{3, 3 + 5, 3 + 6, 3 + 9, 3 + 15\} = \{3, 8, 9, 12, 18\}
   \]

3. **Calculate the mean**: The mean is the sum of the numbers divided by the number of elements. First, we find the sum:
   \[
   3 + 8 + 9 + 12 + 18
   \]
   Adding these step-by-step:
   \[
   3 + 8 = 11
   \]
   \[
   11 + 9 = 20
   \]
   \[
   20 + 12 = 32
   \]
   \[
   32 + 18 = 50
   \]
   The sum of the numbers is 50. Since there are 5 numbers, the mean is:
   \[
   \frac{50}{5} = 10
   \]

4. **Alternative method to find the sum**: We can also find the sum by noting that the set is \(\{n, n+5, n+6, n+9, n+15\}\). The sum of the terms is:
   \[
   n + (n + 5) + (n + 6) + (n + 9) + (n + 15) = 5n + 35
   \]
   Substituting \(n = 3\):
   \[
   5(3) + 35 = 15 + 35 = 50
   \]
   The mean is then:
   \[
   \frac{50}{5} = 10
   \]

Therefore, the mean of the set is \(\boxed{10}\).

System: You are a helpful assistant.


[User] Find the derivative of \( y = \sqrt{2} e^{\sqrt{2} x} \).

[Assistant]
To find the derivative of \( y = \sqrt{2} e^{\sqrt{2} x} \), we proceed as follows:

1. **Identify the function and its components**: The function is \( y = \sqrt{2} e^{\sqrt{2} x} \). Here, \(\sqrt{2}\) is a constant coefficient, and \( e^{\sqrt{2} x} \) is an exponential function.

2. **Apply the constant multiple rule**: The derivative of a constant times a function is the constant times the derivative of the function. Therefore,
   \[
   \frac{d}{dx} \left( \sqrt{2} e^{\sqrt{2} x} \right) = \sqrt{2} \cdot \frac{d}{dx} \left( e^{\sqrt{2} x} \right).
   \]

3. **Differentiate the exponential function using the chain rule**: The chain rule states that the derivative of \( e^{u(x)} \) is \( e^{u(x)} \cdot u'(x) \). Here, \( u(x) = \sqrt{2} x \), so \( u'(x) = \sqrt{2} \). Therefore,
   \[
   \frac{d}{dx} \left( e^{\sqrt{2} x} \right) = e^{\sqrt{2} x} \cdot \sqrt{2}.
   \]

4. **Combine the results**: Substitute the derivative of the exponential function back into the constant multiple rule:
   \[
   \frac{d}{dx} \left( \sqrt{2} e^{\sqrt{2} x} \right) = \sqrt{2} \cdot \left( e^{\sqrt{2} x} \cdot \sqrt{2} \right).
   \]

5. **Simplify the expression**: Multiply the constants \(\sqrt{2}\) and \(\sqrt{2}\):
   \[
   \sqrt{2} \cdot \sqrt{2} = 2.
   \]
   Therefore,
   \[
   \frac{d}{dx} \left( \sqrt{2} e^{\sqrt{2} x} \right) = 2 e^{\sqrt{2} x}.
   \]

The final answer is:
\[
\boxed{2 e^{\sqrt{2} x}}
\]

System: You are a helpful assistant.


[User] How many distinguishable ways are there to construct a regular octahedron using eight congruent equilateral triangles, each of a different color? Two colored octahedrons are considered distinguishable if neither can be rotated to look just like the other.

[Assistant]
To determine the number of distinguishable ways to construct a regular octahedron using eight congruent equilateral triangles, each of a different color, we need to account for the rotational symmetries of the octahedron. The rotational symmetry group of a regular octahedron (which is isomorphic to the rotational symmetry group of a cube) has 24 elements.

Given that each of the eight faces is a different color, we start by calculating the total number of colorings without considering symmetry. This is simply the number of permutations of the 8 colors, which is \(8!\):

\[
8! = 40320
\]

Next, we apply Burnside's Lemma to count the number of distinct colorings under the group of rotational symmetries. Burnside's Lemma states that the number of distinct colorings is the average number of colorings fixed by each group element. Mathematically, this is expressed as:

\[
\text{Number of distinct colorings} = \frac{1}{|G|} \sum_{g \in G} \text{fix}(g)
\]

where \(|G|\) is the order of the group (24 in this case), and \(\text{fix}(g)\) is the number of colorings fixed by the group element \(g\).

For a non-identity rotation \(g\), the coloring must remain unchanged under the rotation. However, since all faces are distinct colors, any non-identity rotation that permutes the faces would require those faces to have the same color, which is impossible. Therefore, \(\text{fix}(g) = 0\) for all non-identity rotations.

The only rotation that fixes any coloring is the identity rotation, which leaves all faces in place. For the identity rotation, every coloring is fixed, so \(\text{fix}(\text{id}) = 8!\).

Thus, the sum \(\sum_{g \in G} \text{fix}(g)\) is simply \(8!\) because the identity rotation is the only one that fixes any coloring:

\[
\sum_{g \in G} \text{fix}(g) = 8!
\]

Applying Burnside's Lemma, we get:

\[
\text{Number of distinct colorings} = \frac{1}{24} \times 8! = \frac{40320}{24} = 1680
\]

Therefore, the number of distinguishable ways to construct the octahedron is:

\[
\boxed{1680}
\]

System: You are a helpful assistant.


[User]
Calculate the integral
$$\int\limits_{0}^{1}dx\int\limits_{x}^{1}dy\int\limits_{y}^{1}e^{z^3}dz$$

[Assistant]
To solve the given triple integral
\[
\int_{0}^{1} \int_{x}^{1} \int_{y}^{1} e^{z^3} \, dz \, dy \, dx,
\]
we will change the order of integration to simplify the computation.

### Step 1: Change the Order of Integration
The original region of integration is defined by \(0 \le x \le y \le z \le 1\). To change the order of integration to \(dx \, dy \, dz\), we need to express the same region in terms of \(z\), \(y\), and \(x\).

For a fixed \(z\):
- \(y\) ranges from \(0\) to \(z\).
- For each \(y\), \(x\) ranges from \(0\) to \(y\).

Thus, the integral becomes:
\[
\int_{0}^{1} \int_{0}^{z} \int_{0}^{y} e^{z^3} \, dx \, dy \, dz.
\]

### Step 2: Integrate with Respect to \(x\)
First, integrate the innermost integral with respect to \(x\):
\[
\int_{0}^{y} dx = y.
\]
So the integral now is:
\[
\int_{0}^{1} \int_{0}^{z} y e^{z^3} \, dy \, dz.
\]

### Step 3: Integrate with Respect to \(y\)
Next, integrate with respect to \(y\):
\[
\int_{0}^{z} y \, dy = \left[ \frac{1}{2} y^2 \right]_{0}^{z} = \frac{1}{2} z^2.
\]
So the integral now is:
\[
\int_{0}^{1} \frac{1}{2} z^2 e^{z^3} \, dz.
\]

### Step 4: Integrate with Respect to \(z\) Using Substitution
Finally, integrate with respect to \(z\). Use the substitution \(u = z^3\), so \(du = 3z^2 \, dz\), or \(z^2 \, dz = \frac{1}{3} du\).

When \(z = 0\), \(u = 0\); when \(z = 1\), \(u = 1\). The integral becomes:
\[
\frac{1}{2} \int_{0}^{1} z^2 e^{z^3} \, dz = \frac{1}{2} \cdot \frac{1}{3} \int_{0}^{1} e^u \, du = \frac{1}{6} \left[ e^u \right]_{0}^{1} = \frac{1}{6} (e - 1).
\]

### Final Answer
The value of the integral is:
\[
\boxed{\frac{e - 1}{6}}.
\]

System: You are a helpful assistant.


[User] A very large number $x$ is equal to $2^23^34^45^56^67^78^89^9$. What is the smallest positive integer that, when multiplied with $x$, produces a product that is a perfect square?

[Assistant]
To find the smallest positive integer that, when multiplied with \( x = 2^2 \times 3^3 \times 4^4 \times 5^5 \times 6^6 \times 7^7 \times 8^8 \times 9^9 \), produces a perfect square, we start by expressing \( x \) in terms of its prime factors.

First, we factorize each term:
- \( 4^4 = (2^2)^4 = 2^8 \)
- \( 6^6 = (2 \times 3)^6 = 2^6 \times 3^6 \)
- \( 8^8 = (2^3)^8 = 2^{24} \)
- \( 9^9 = (3^2)^9 = 3^{18} \)

Now, we combine all the prime factors:
\[ x = 2^2 \times 3^3 \times 2^8 \times 5^5 \times 2^6 \times 3^6 \times 7^7 \times 2^{24} \times 3^{18} \]

Next, we sum the exponents for each prime factor:
- For \( 2 \): \( 2 + 8 + 6 + 24 = 40 \)
- For \( 3 \): \( 3 + 6 + 18 = 27 \)
- For \( 5 \): \( 5 \)
- For \( 7 \): \( 7 \)

Thus, the prime factorization of \( x \) is:
\[ x = 2^{40} \times 3^{27} \times 5^5 \times 7^7 \]

For \( x \) to be a perfect square, all the exponents in its prime factorization must be even. We check each exponent:
- \( 2^{40} \): The exponent 40 is already even.
- \( 3^{27} \): The exponent 27 is odd, so we need to multiply by \( 3^1 \) to make it even.
- \( 5^5 \): The exponent 5 is odd, so we need to multiply by \( 5^1 \) to make it even.
- \( 7^7 \): The exponent 7 is odd, so we need to multiply by \( 7^1 \) to make it even.

Therefore, the smallest positive integer that, when multiplied with \( x \), makes \( x \) a perfect square is:
\[ 3^1 \times 5^1 \times 7^1 = 3 \times 5 \times 7 = 105 \]

Thus, the smallest positive integer is:
\[
\boxed{105}
\]

System: You are a helpful assistant.


[User] If \( x \) and \( y \) are related and \( \frac{d^2x}{dy^2} \cdot \left( \frac{dy}{dx} \right)^3 + K \frac{d^2y}{dx^2} = 0 \), then what is the value of \( K \)?

[Assistant]
To find the value of \( K \) in the equation

\[
\frac{d^2x}{dy^2} \cdot \left( \frac{dy}{dx} \right)^3 + K \frac{d^2y}{dx^2} = 0,
\]

we start by expressing \(\frac{d^2x}{dy^2}\) in terms of derivatives of \( y \) with respect to \( x \).

First, recall that if \( y \) is a function of \( x \), then the first derivative of \( x \) with respect to \( y \) is

\[
\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}.
\]

Next, we need to find the second derivative \(\frac{d^2x}{dy^2}\). Using the chain rule, we have

\[
\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dx} \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{dx}{dy}.
\]

Calculating the derivative with respect to \( x \):

\[
\frac{d}{dx} \left( \frac{1}{\frac{dy}{dx}} \right) = -\frac{\frac{d^2y}{dx^2}}{\left( \frac{dy}{dx} \right)^2}.
\]

Multiplying by \(\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}\):

\[
\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{\left( \frac{dy}{dx} \right)^3}.
\]

Now, substitute this expression into the original equation:

\[
-\frac{\frac{d^2y}{dx^2}}{\left( \frac{dy}{dx} \right)^3} \cdot \left( \frac{dy}{dx} \right)^3 + K \frac{d^2y}{dx^2} = 0.
\]

Simplify the first term:

\[
-\frac{d^2y}{dx^2} + K \frac{d^2y}{dx^2} = 0.
\]

Factor out \(\frac{d^2y}{dx^2}\):

\[
(-1 + K) \frac{d^2y}{dx^2} = 0.
\]

Since this equation must hold for any function \( y(x) \), the coefficient must be zero:

\[
-1 + K = 0 \implies K = 1.
\]

Thus, the value of \( K \) is \(\boxed{1}\).

System: You are a helpful assistant.


[User] What is the area enclosed by the graph of $|3x|+|4y|=12$?

[Assistant]
To find the area enclosed by the graph of \(|3x| + |4y| = 12\), we first determine the vertices of the shape formed by this equation.

By considering the different cases for the signs of \(x\) and \(y\):

1. In the first quadrant (\(x \geq 0\), \(y \geq 0\)):
   \[
   3x + 4y = 12
   \]
   When \(x = 0\), \(y = 3\). When \(y = 0\), \(x = 4\).

2. In the second quadrant (\(x \leq 0\), \(y \geq 0\)):
   \[
   -3x + 4y = 12
   \]
   When \(x = 0\), \(y = 3\). When \(y = 0\), \(x = -4\).

3. In the third quadrant (\(x \leq 0\), \(y \leq 0\)):
   \[
   -3x - 4y = 12
   \]
   When \(x = 0\), \(y = -3\). When \(y = 0\), \(x = -4\).

4. In the fourth quadrant (\(x \geq 0\), \(y \leq 0\)):
   \[
   3x - 4y = 12
   \]
   When \(x = 0\), \(y = -3\). When \(y = 0\), \(x = 4\).

These calculations show that the vertices of the shape are \((4, 0)\), \((-4, 0)\), \((0, 3)\), and \((0, -3)\). This shape is a rhombus centered at the origin with horizontal vertices at \((\pm 4, 0)\) and vertical vertices at \((0, \pm 3)\).

To find the area of the rhombus, we use the formula for the area of a rhombus, which is given by:
\[
\text{Area} = \frac{1}{2} \times \text{product of the lengths of the diagonals}
\]

The length of the horizontal diagonal is \(8\) (from \((-4, 0)\) to \((4, 0)\)), and the length of the vertical diagonal is \(6\) (from \((0, -3)\) to \((0, 3)\)).

Thus, the area is:
\[
\text{Area} = \frac{1}{2} \times 8 \times 6 = 24
\]

Therefore, the area enclosed by the graph \(|3x| + |4y| = 12\) is \(\boxed{24}\).

System: You are a helpful assistant.


[User] Evaluate three level surfaces of the function \( f(x,y,z) = \sqrt{x^2 + y^2 + z^2} \).

[Assistant]
To evaluate three level surfaces of the function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), we start by recognizing that the level surfaces are defined by the equation \( f(x, y, z) = c \), where \( c \) is a constant.

Given \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), we set this equal to \( c \):
\[
\sqrt{x^2 + y^2 + z^2} = c
\]

Squaring both sides, we obtain:
\[
x^2 + y^2 + z^2 = c^2
\]

This equation represents a sphere centered at the origin with radius \( c \).

To evaluate three specific level surfaces, we choose three different values for \( c \). Let's take \( c = 1 \), \( c = 2 \), and \( c = 3 \).

For \( c = 1 \):
\[
x^2 + y^2 + z^2 = 1
\]

For \( c = 2 \):
\[
x^2 + y^2 + z^2 = 4
\]

For \( c = 3 \):
\[
x^2 + y^2 + z^2 = 9
\]

Thus, the three level surfaces are the spheres given by:
\[
\boxed{x^2 + y^2 + z^2 = 1}, \boxed{x^2 + y^2 + z^2 = 4}, \boxed{x^2 + y^2 + z^2 = 9}
\]

System: You are a helpful assistant.


[User] Forest Ho-Chen notices that the stars on the US flag are arranged as a $6$ by $5$ rectangle with a $5$ by $4$ rectangle on the inside, allowing for beautiful cross patterns. He wonders what the previous number of states the US could have had where the stars could be arranged in a $x + 1$ by $x$ rectangle with an $x$ by $x - 1$ rectangle on the inside.

[Assistant]
To solve the problem, we need to determine the number of states the US could have had where the stars can be arranged in a specific pattern. The pattern is described as an \((x+1) \times x\) rectangle with an \(x \times (x-1)\) rectangle on the inside.

First, we need to derive the formula for the total number of stars. The total number of stars is the sum of the stars in the outer rectangle and the inner rectangle. The outer rectangle has \((x+1) \times x\) stars, and the inner rectangle has \(x \times (x-1)\) stars. Therefore, the total number of stars \(N\) is given by:
\[
N = (x+1)x + x(x-1)
\]

Simplifying the expression:
\[
N = x(x+1) + x(x-1) = x^2 + x + x^2 - x = 2x^2
\]

So, the number of stars is \(2x^2\).

Given that the current number of stars is 50, we can find the value of \(x\) for the current flag:
\[
2x^2 = 50 \implies x^2 = 25 \implies x = 5
\]

To find the previous number of states, we need to find the largest \(x\) such that \(2x^2 < 50\). We test the values of \(x\) less than 5:

- For \(x = 4\):
  \[
  2 \times 4^2 = 2 \times 16 = 32
  \]

- For \(x = 3\):
  \[
  2 \times 3^2 = 2 \times 9 = 18
  \]

- For \(x = 2\):
  \[
  2 \times 2^2 = 2 \times 4 = 8
  \]

- For \(x = 1\):
  \[
  2 \times 1^2 = 2 \times 1 = 2
  \]

The largest value of \(2x^2\) that is less than 50 is 32, which corresponds to \(x = 4\).

Therefore, the previous number of states the US could have had is \(\boxed{32}\).

System: You are a helpful assistant.


[User] When the square root of $x$ is cubed, the answer is 64. What is the value of $x$?

[Assistant]
To solve the problem, we start with the equation given in the problem statement:

\[
(\sqrt{x})^3 = 64
\]

We can rewrite the square root as an exponent:

\[
(x^{1/2})^3 = 64
\]

Using the property of exponents \((a^m)^n = a^{mn}\), we simplify the left side:

\[
x^{3/2} = 64
\]

To isolate \(x\), we raise both sides of the equation to the power of \(\frac{2}{3}\):

\[
(x^{3/2})^{2/3} = 64^{2/3}
\]

Simplifying the left side, we get:

\[
x = 64^{2/3}
\]

Next, we need to compute \(64^{2/3}\). We can break this down into two steps:

1. Compute the cube root of 64:

\[
64^{1/3} = 4
\]

2. Square the result:

\[
4^2 = 16
\]

Thus, we have:

\[
x = 16
\]

To verify, we substitute \(x = 16\) back into the original equation:

\[
(\sqrt{16})^3 = 4^3 = 64
\]

This confirms that our solution is correct. Therefore, the value of \(x\) is:

\[
\boxed{16}
\]

System: You are a helpful assistant.


[User] Integrate $\frac{1}{1-x^{2}}$ using the substitution $x = \tanh z$.

[Assistant]
To integrate \(\frac{1}{1 - x^2}\) using the substitution \(x = \tanh z\), follow these steps:

1. **Substitution**: Let \(x = \tanh z\). Then, the differential \(dx\) can be expressed as:
   \[
   dx = \text{sech}^2 z \, dz
   \]

2. **Rewrite the integrand**: Using the identity \(1 - \tanh^2 z = \text{sech}^2 z\), we have:
   \[
   \frac{1}{1 - x^2} = \frac{1}{1 - \tanh^2 z} = \frac{1}{\text{sech}^2 z} = \cosh^2 z
   \]

3. **Substitute into the integral**: Substitute \(x = \tanh z\) and \(dx = \text{sech}^2 z \, dz\) into the integral:
   \[
   \int \frac{1}{1 - x^2} \, dx = \int \cosh^2 z \cdot \text{sech}^2 z \, dz
   \]

4. **Simplify the integrand**: Notice that \(\cosh^2 z \cdot \text{sech}^2 z = 1\), so the integral becomes:
   \[
   \int 1 \, dz
   \]

5. **Integrate**: The integral of 1 with respect to \(z\) is:
   \[
   \int 1 \, dz = z + C
   \]

6. **Back-substitute**: Recall that \(z = \tanh^{-1} x\), so:
   \[
   z + C = \tanh^{-1} x + C
   \]

Therefore, the integral of \(\frac{1}{1 - x^2}\) using the substitution \(x = \tanh z\) is:
\[
\boxed{\tanh^{-1} x + C}
\]

System: You are a helpful assistant.


[User] A giant, circular and perfectly flat pizza must be shared by 22 people. What is the least number of cuts to be made on the pizza so that each person can have a piece of pizza? (The pieces are not necessarily the same shape or with the same area.)

[Assistant]
To determine the minimum number of cuts required to divide a circular pizza into at least 22 pieces, we use the formula for the maximum number of regions \( R(n) \) created by \( n \) straight cuts, which is given by:

\[
R(n) = \frac{n^2 + n + 2}{2}
\]

We need to find the smallest \( n \) such that \( R(n) \geq 22 \). Therefore, we solve the inequality:

\[
\frac{n^2 + n + 2}{2} \geq 22
\]

First, multiply both sides by 2 to clear the fraction:

\[
n^2 + n + 2 \geq 44
\]

Next, subtract 44 from both sides to set the inequality to zero:

\[
n^2 + n + 2 - 44 \geq 0
\]

Simplify the equation:

\[
n^2 + n - 42 \geq 0
\]

To solve this quadratic inequality, we first find the roots of the quadratic equation \( n^2 + n - 42 = 0 \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -42 \):

\[
n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-42)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm \sqrt{169}}{2} = \frac{-1 \pm 13}{2}
\]

This gives us two solutions:

\[
n = \frac{12}{2} = 6 \quad \text{and} \quad n = \frac{-14}{2} = -7
\]

Since \( n \) must be a non-negative integer, we discard \( n = -7 \) and keep \( n = 6 \).

We verify that \( n = 6 \) satisfies the inequality:

\[
R(6) = \frac{6^2 + 6 + 2}{2} = \frac{36 + 6 + 2}{2} = \frac{44}{2} = 22
\]

Thus, with 6 cuts, we can create exactly 22 regions, which is sufficient to serve 22 people.

Therefore, the minimum number of cuts required is:

\[
\boxed{6}
\]

System: You are a helpful assistant.


[User] If $2^{10} \cdot 2^{15}$ is expressed as some integer to the fifth power, what is that integer?

[Assistant]
To solve the problem of expressing \(2^{10} \cdot 2^{15}\) as an integer to the fifth power, we start by simplifying the expression using the properties of exponents. Specifically, when multiplying exponents with the same base, we add the exponents:

\[
2^{10} \cdot 2^{15} = 2^{10 + 15} = 2^{25}
\]

Next, we need to express \(2^{25}\) as an integer raised to the fifth power. We can rewrite \(2^{25}\) as:

\[
2^{25} = (2^5)^5
\]

Now, we calculate \(2^5\):

\[
2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32
\]

Therefore, we have:

\[
2^{25} = 32^5
\]

Thus, the integer that, when raised to the fifth power, equals \(2^{25}\) is:

\[
\boxed{32}
\]

System: You are a helpful assistant.


[User] How many prime numbers are between 30 and 40?

[Assistant]
To determine how many prime numbers are between 30 and 40, we first list the numbers in this range: 31, 32, 33, 34, 35, 36, 37, 38, 39.

A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. We will check each number for primality by testing divisibility by primes less than or equal to the square root of the number.

1. **31**: The square root of 31 is approximately 5.5, so we check divisibility by primes up to 5 (i.e., 2, 3, and 5).
   - 31 is not divisible by 2 (31 ÷ 2 = 15.5, not an integer).
   - 31 is not divisible by 3 (31 ÷ 3 ≈ 10.33, not an integer).
   - 31 is not divisible by 5 (31 ÷ 5 = 6.2, not an integer).
   - Therefore, 31 is a prime number.

2. **32**: 32 is even, so it is divisible by 2. Therefore, 32 is not a prime number.

3. **33**: 33 is divisible by 3 (33 ÷ 3 = 11). Therefore, 33 is not a prime number.

4. **34**: 34 is even, so it is divisible by 2. Therefore, 34 is not a prime number.

5. **35**: 35 ends in 5, so it is divisible by 5 (35 ÷ 5 = 7). Therefore, 35 is not a prime number.

6. **36**: 36 is even, so it is divisible by 2. Therefore, 36 is not a prime number.

7. **37**: The square root of 37 is approximately 6.08, so we check divisibility by primes up to 5 (i.e., 2, 3, and 5).
   - 37 is not divisible by 2 (37 ÷ 2 = 18.5, not an integer).
   - 37 is not divisible by 3 (37 ÷ 3 ≈ 12.33, not an integer).
   - 37 is not divisible by 5 (37 ÷ 5 = 7.4, not an integer).
   - Therefore, 37 is a prime number.

8. **38**: 38 is even, so it is divisible by 2. Therefore, 38 is not a prime number.

9. **39**: 39 is divisible by 3 (39 ÷ 3 = 13). Therefore, 39 is not a prime number.

Thus, the prime numbers between 30 and 40 are 31 and 37. Therefore, there are 2 prime numbers in this range.

\[
\boxed{2}
\]

System: You are a helpful assistant.


[User] Find a general solution for \( n \) such that \( x^n + x - 1 \equiv 0 \pmod{x^2 - x + 1} \).

[Assistant]
To solve the problem of finding all integers \( n \) such that \( x^n + x - 1 \equiv 0 \pmod{x^2 - x + 1} \), we start by noting that we need \( x^n \equiv -x + 1 \pmod{x^2 - x + 1} \).

First, we reduce higher powers of \( x \) modulo \( x^2 - x + 1 \) using the relation \( x^2 \equiv x - 1 \pmod{x^2 - x + 1} \). We calculate the powers of \( x \) modulo \( x^2 - x + 1 \):

\[
\begin{aligned}
x^0 &\equiv 1 \pmod{x^2 - x + 1}, \\
x^1 &\equiv x \pmod{x^2 - x + 1}, \\
x^2 &\equiv x - 1 \pmod{x^2 - x + 1}, \\
x^3 &\equiv x \cdot x^2 \equiv x \cdot (x - 1) \equiv x^2 - x \equiv (x - 1) - x \equiv -1 \pmod{x^2 - x + 1}, \\
x^4 &\equiv x \cdot x^3 \equiv x \cdot (-1) \equiv -x \pmod{x^2 - x + 1}, \\
x^5 &\equiv x \cdot x^4 \equiv x \cdot (-x) \equiv -x^2 \equiv -(x - 1) \equiv -x + 1 \pmod{x^2 - x + 1}, \\
x^6 &\equiv x \cdot x^5 \equiv x \cdot (-x + 1) \equiv -x^2 + x \equiv -(x - 1) + x \equiv 1 \pmod{x^2 - x + 1}.
\end{aligned}
\]

We observe that the powers of \( x \) repeat every 6 steps. Therefore, \( x^n \equiv x^{n \mod 6} \pmod{x^2 - x + 1} \).

From the calculations, we see that \( x^5 \equiv -x + 1 \pmod{x^2 - x + 1} \). Substituting this into the original equation, we get:

\[
x^5 + x - 1 \equiv (-x + 1) + x - 1 \equiv 0 \pmod{x^2 - x + 1}.
\]

Thus, \( n \) must satisfy \( n \equiv 5 \pmod{6} \). The general solution is \( n = 6k + 5 \) for integers \( k \).

Therefore, the general solution for \( n \) is \(\boxed{n \equiv 5 \pmod{6}}\).

System: You are a helpful assistant.


[User] In a group of 50 people, 36 like dogs and 25 like cats. Let \( x \) denote the number of people who like both dogs and cats. How many different values can \( x \) take?

[Assistant]
To determine the number of different values \( x \) (the number of people who like both dogs and cats) can take in a group of 50 people where 36 like dogs and 25 like cats, we use the principle of inclusion-exclusion.

The number of people who like at least one of the pets (dogs or cats) is given by:
\[
36 + 25 - x = 61 - x
\]

Since the total number of people in the group is 50, the number of people who like at least one pet must be less than or equal to 50:
\[
61 - x \leq 50
\]
Solving for \( x \):
\[
61 - x \leq 50 \implies x \geq 11
\]

The maximum value of \( x \) is constrained by the smaller of the two groups (dogs or cats). Since 25 people like cats, the maximum value of \( x \) is 25:
\[
x \leq 25
\]

Therefore, \( x \) must satisfy:
\[
11 \leq x \leq 25
\]

The integer values of \( x \) in this range are from 11 to 25 inclusive. The number of integers in this range is:
\[
25 - 11 + 1 = 15
\]

Thus, the number of different values \( x \) can take is \(\boxed{15}\).

System: You are a helpful assistant.