[User] How do you factor $x^n - 1$, where $n \geq 3$, into irreducibles in $\mathbb{Z}[x]$?

[Assistant]
To factor the polynomial \( x^n - 1 \) into irreducibles in \( \mathbb{Z}[x] \) for \( n \geq 3 \), we use the concept of cyclotomic polynomials. The \( n \)-th cyclotomic polynomial, denoted \( \Phi_n(x) \), is the minimal polynomial over the integers for the primitive \( n \)-th roots of unity. The polynomial \( x^n - 1 \) can be factored as the product of cyclotomic polynomials corresponding to the divisors of \( n \).

The factorization is given by:

\[
x^n - 1 = \prod_{d \mid n} \Phi_d(x),
\]

where the product is taken over all positive divisors \( d \) of \( n \). Each \( \Phi_d(x) \) is irreducible over \( \mathbb{Z}[x] \).

To verify this, consider the following steps:

1. **Definition of Cyclotomic Polynomials**: The \( d \)-th cyclotomic polynomial \( \Phi_d(x) \) is defined as the polynomial whose roots are the primitive \( d \)-th roots of unity. These roots are the complex numbers \( e^{2\pi i k/d} \) where \( k \) is coprime to \( d \).

2. **Factorization Property**: The polynomial \( x^n - 1 \) can be written as the product of cyclotomic polynomials for all divisors \( d \) of \( n \). This is a well-known result in algebra and number theory.

3. **Irreducibility of Cyclotomic Polynomials**: Each \( \Phi_d(x) \) is irreducible over \( \mathbb{Z}[x] \). This can be proven using properties of roots of unity and the minimality of the polynomial.

For example, consider \( n = 6 \). The divisors of 6 are 1, 2, 3, and 6. The corresponding cyclotomic polynomials are:

\[
\Phi_1(x) = x - 1,
\]
\[
\Phi_2(x) = x + 1,
\]
\[
\Phi_3(x) = x^2 + x + 1,
\]
\[
\Phi_6(x) = x^2 - x + 1.
\]

Thus, the factorization of \( x^6 - 1 \) is:

\[
x^6 - 1 = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1).
\]

Each of these polynomials is irreducible over \( \mathbb{Z}[x] \).

Therefore, the factorization of \( x^n - 1 \) into irreducibles in \( \mathbb{Z}[x] \) is:

\[
\boxed{x^n - 1 = \prod_{d \mid n} \Phi_d(x)}.
\]

System: You are a helpful assistant.


[User] Suppose that the number of people entering a department store in a given day is a random variable with mean 50. Suppose further that the amount of money spent by these customers are random variables with a common mean of 8. Assume also that the amount of money spent by a customer is independent of the total number of customers that have entered the store. What is the expected amount of money spent in the store on a given day?

[Assistant]
To find the expected amount of money spent in the store on a given day, we need to consider the random variables involved. Let \( N \) be the random variable representing the number of people entering the store, with \( E[N] = 50 \). Let \( X_i \) be the amount of money spent by the \( i \)-th customer, with \( E[X_i] = 8 \) for all \( i \).

The total amount of money spent in the store on a given day is given by the sum of the amounts spent by each customer:
\[ T = \sum_{i=1}^N X_i \]

To find the expected value of \( T \), we use the law of total expectation:
\[ E[T] = E\left[ \sum_{i=1}^N X_i \right] \]

By the linearity of expectation, we have:
\[ E\left[ \sum_{i=1}^N X_i \right] = E\left[ E\left[ \sum_{i=1}^N X_i \mid N \right] \right] \]

Given \( N \), the expected value of the sum is:
\[ E\left[ \sum_{i=1}^N X_i \mid N \right] = N \cdot E[X_i] \]

Since \( E[X_i] = 8 \), we get:
\[ E\left[ \sum_{i=1}^N X_i \mid N \right] = N \cdot 8 \]

Taking the expectation over \( N \), we obtain:
\[ E[T] = E[N \cdot 8] = 8 \cdot E[N] \]

Given that \( E[N] = 50 \), we have:
\[ E[T] = 8 \cdot 50 = 400 \]

Therefore, the expected amount of money spent in the store on a given day is:
\[ \boxed{400} \]

System: You are a helpful assistant.


[User] A square has an area of $25x^2 + 20x + 4$. Find the restrictions on $x$ such that the side length of the square is positive.

[Assistant]
To find the restrictions on \( x \) such that the side length of the square is positive, given that the area of the square is \( 25x^2 + 20x + 4 \), we can proceed as follows:

1. **Identify the Side Length:**
   The area of a square is given by the square of its side length. Therefore, if the area is \( 25x^2 + 20x + 4 \), we need to determine if this expression can be written as a perfect square.

2. **Check for Perfect Square:**
   We observe that \( 25x^2 + 20x + 4 \) can be factored as a perfect square trinomial:
   \[
   25x^2 + 20x + 4 = (5x + 2)^2
   \]
   This means the side length of the square is \( 5x + 2 \).

3. **Ensure Positive Side Length:**
   For the side length to be positive, we need:
   \[
   5x + 2 > 0
   \]

4. **Solve the Inequality:**
   Solving the inequality \( 5x + 2 > 0 \):
   \[
   5x + 2 > 0 \implies 5x > -2 \implies x > -\frac{2}{5}
   \]

5. **Conclusion:**
   The restriction on \( x \) such that the side length of the square is positive is:
   \[
   \boxed{x > -\frac{2}{5}}
   \]

System: You are a helpful assistant.


[User] If real numbers $x$ and $y$ satisfy $(x+2y)^2 + 4(x+2y + 2 - xy) = 0,$ what is $x+2y?$

[Assistant]
To solve the equation \((x + 2y)^2 + 4(x + 2y + 2 - xy) = 0\) and find the value of \(x + 2y\), we proceed as follows:

First, let \(t = x + 2y\). The equation then becomes:
\[
t^2 + 4(t + 2 - xy) = 0
\]

Next, we expand and simplify the original equation directly:
\[
(x + 2y)^2 + 4(x + 2y + 2 - xy) = x^2 + 4xy + 4y^2 + 4x + 8y + 8 - 4xy
\]
The \(4xy\) and \(-4xy\) terms cancel out, leaving:
\[
x^2 + 4y^2 + 4x + 8y + 8 = 0
\]

We complete the square for the \(x\) and \(y\) terms:
\[
x^2 + 4x = (x + 2)^2 - 4
\]
\[
4y^2 + 8y = 4(y^2 + 2y) = 4[(y + 1)^2 - 1] = 4(y + 1)^2 - 4
\]

Substituting these into the equation, we get:
\[
(x + 2)^2 - 4 + 4(y + 1)^2 - 4 + 8 = 0
\]
Simplifying the constants:
\[
(x + 2)^2 + 4(y + 1)^2 = 0
\]

Since the sum of squares is zero, each term must be zero:
\[
(x + 2)^2 = 0 \quad \text{and} \quad 4(y + 1)^2 = 0
\]
Thus:
\[
x + 2 = 0 \quad \text{and} \quad y + 1 = 0
\]
So:
\[
x = -2 \quad \text{and} \quad y = -1
\]

Substituting \(x = -2\) and \(y = -1\) back into \(t = x + 2y\):
\[
t = -2 + 2(-1) = -2 - 2 = -4
\]

Therefore, the value of \(x + 2y\) is \(\boxed{-4}\).

System: You are a helpful assistant.


[User] How many vertical asymptotes does the function \( f(x) = \frac{x^2 - 6x - 16}{x^3 - 5x^2 - 2x + 24} \) have?

[Assistant]
To determine the number of vertical asymptotes of the function \( f(x) = \frac{x^2 - 6x - 16}{x^3 - 5x^2 - 2x + 24} \), we need to factor both the numerator and the denominator and identify the points where the denominator is zero but the numerator is not zero.

1. **Factor the numerator:**
   \[
   x^2 - 6x - 16 = (x - 8)(x + 2)
   \]
   The roots of the numerator are \( x = 8 \) and \( x = -2 \).

2. **Factor the denominator:**
   We use the Rational Root Theorem to find the roots of the cubic polynomial \( x^3 - 5x^2 - 2x + 24 \). Testing possible rational roots, we find:
   \[
   x = 3 \quad \text{(since } 3^3 - 5 \cdot 3^2 - 2 \cdot 3 + 24 = 0 \text{)}
   \]
   Using synthetic division or polynomial division, we can factor \( x^3 - 5x^2 - 2x + 24 \) as:
   \[
   x^3 - 5x^2 - 2x + 24 = (x - 3)(x^2 - 2x - 8)
   \]
   Further factoring the quadratic:
   \[
   x^2 - 2x - 8 = (x - 4)(x + 2)
   \]
   Thus, the denominator factors to:
   \[
   x^3 - 5x^2 - 2x + 24 = (x - 3)(x - 4)(x + 2)
   \]
   The roots of the denominator are \( x = 3 \), \( x = 4 \), and \( x = -2 \).

3. **Simplify the function:**
   \[
   f(x) = \frac{(x - 8)(x + 2)}{(x - 3)(x - 4)(x + 2)}
   \]
   We can cancel the common factor \( (x + 2) \) from the numerator and the denominator, provided \( x \neq -2 \):
   \[
   f(x) = \frac{x - 8}{(x - 3)(x - 4)}, \quad \text{for } x \neq -2
   \]

4. **Identify vertical asymptotes:**
   Vertical asymptotes occur where the denominator of the simplified function is zero, but the numerator is not zero. The simplified denominator \( (x - 3)(x - 4) \) is zero at \( x = 3 \) and \( x = 4 \). At these points, the numerator \( (x - 8) \) is not zero.

   The factor \( x + 2 \) was canceled, indicating a hole at \( x = -2 \), not a vertical asymptote.

Therefore, the vertical asymptotes are at \( x = 3 \) and \( x = 4 \).

\[
\boxed{2}
\]

System: You are a helpful assistant.


[User] Solve the equation \( e^{1-x} = 17.5 \).

[Assistant]
To solve the equation \( e^{1-x} = 17.5 \), we start by taking the natural logarithm of both sides of the equation. This gives:

\[
\ln(e^{1-x}) = \ln(17.5)
\]

Since the natural logarithm and the exponential function are inverses, we can simplify the left side:

\[
1 - x = \ln(17.5)
\]

Next, we solve for \( x \) by isolating it on one side of the equation. Subtract \( \ln(17.5) \) from both sides:

\[
1 - \ln(17.5) = x
\]

Therefore, the solution to the equation is:

\[
x = 1 - \ln(17.5)
\]

To verify, we substitute \( x = 1 - \ln(17.5) \) back into the original equation:

\[
e^{1 - (1 - \ln(17.5))} = e^{\ln(17.5)} = 17.5
\]

This confirms that the solution is correct. Thus, the final answer is:

\[
\boxed{1 - \ln(17.5)}
\]

System: You are a helpful assistant.


[User] Compute the limit of the sequence $\{z_n\}$ where $z_n = i^{n!}$.

System: You are a helpful assistant.


[User] Given matrices $\boldsymbol{A}=\begin{bmatrix} 1 & 3 & -4 \\ -1 & -3 & 4 \end{bmatrix}$ and $\boldsymbol{B}=\begin{bmatrix} 1 & 3 \\ -1 & 8 \end{bmatrix}$, find the product $AB$.

[Assistant]
To find the product \( \boldsymbol{AB} \), we need to check if the dimensions of the matrices allow for multiplication. Specifically, the number of columns in matrix \( \boldsymbol{A} \) must equal the number of rows in matrix \( \boldsymbol{B} \).

Given:
\[ \boldsymbol{A} = \begin{bmatrix} 1 & 3 & -4 \\ -1 & -3 & 4 \end{bmatrix} \]
\[ \boldsymbol{B} = \begin{bmatrix} 1 & 3 \\ -1 & 8 \end{bmatrix} \]

Matrix \( \boldsymbol{A} \) is a \( 2 \times 3 \) matrix (2 rows and 3 columns).
Matrix \( \boldsymbol{B} \) is a \( 2 \times 2 \) matrix (2 rows and 2 columns).

For the product \( \boldsymbol{AB} \) to be defined, the number of columns in \( \boldsymbol{A} \) (which is 3) must equal the number of rows in \( \boldsymbol{B} \) (which is 2). Since 3 does not equal 2, the product \( \boldsymbol{AB} \) is undefined.

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

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] Given \( f(t) = t + \ln(t) \), find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \). Provide exact values.

[Assistant]
To find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \) for the function \( f(t) = t + \ln(t) \), we follow these steps:

1. **Verify that \( f \) is invertible:**
   - The domain of \( f \) is \( t > 0 \) because of the \( \ln(t) \) term.
   - Compute the derivative \( f'(t) = 1 + \frac{1}{t} \).
   - Since \( t > 0 \), \( \frac{1}{t} > 0 \), so \( f'(t) = 1 + \frac{1}{t} > 0 \) for all \( t > 0 \).
   - Therefore, \( f \) is strictly increasing and hence invertible.

2. **Find \( f^{-1}(1 + e) \):**
   - Let \( a = f^{-1}(1 + e) \). Then \( f(a) = 1 + e \).
   - This gives the equation \( a + \ln(a) = 1 + e \).
   - Testing \( a = e \):
     \[
     e + \ln(e) = e + 1 = 1 + e
     \]
   - Therefore, \( f^{-1}(1 + e) = e \).

3. **Compute \( f'(e) \):**
   - The derivative of \( f \) is \( f'(t) = 1 + \frac{1}{t} \).
   - Evaluate at \( t = e \):
     \[
     f'(e) = 1 + \frac{1}{e}
     \]

4. **Find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \):**
   - Use the formula for the derivative of the inverse function:
     \[
     \frac{d(f^{-1})}{dt} = \frac{1}{f'(f^{-1}(t))}
     \]
   - Substitute \( t = 1 + e \) and \( f^{-1}(1 + e) = e \):
     \[
     \frac{d(f^{-1})}{dt} \bigg|_{t=1+e} = \frac{1}{f'(e)} = \frac{1}{1 + \frac{1}{e}}
     \]
   - Simplify the expression:
     \[
     \frac{1}{1 + \frac{1}{e}} = \frac{1}{\frac{e + 1}{e}} = \frac{e}{e + 1}
     \]

Thus, the derivative of the inverse function \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \) is:
\[
\boxed{\frac{e}{e + 1}}
\]

System: You are a helpful assistant.


[User] I have a 25% acid solution and a 50% acid solution. How many milliliters of each solution are needed to create a 20 mL solution that is 40% acid?

[Assistant]
To solve the problem, let \( x \) be the amount of the 25% acid solution in milliliters. Consequently, the amount of the 50% acid solution will be \( 20 - x \) milliliters, since the total volume of the final solution is 20 mL.

The equation for the total acid content from both solutions must equal the acid content in the final 40% solution. This can be expressed as:

\[
0.25x + 0.50(20 - x) = 0.40 \times 20
\]

First, simplify the right-hand side:

\[
0.40 \times 20 = 8
\]

So the equation becomes:

\[
0.25x + 0.50(20 - x) = 8
\]

Next, expand and simplify the left-hand side:

\[
0.25x + 0.50 \times 20 - 0.50x = 8
\]
\[
0.25x + 10 - 0.50x = 8
\]
\[
-0.25x + 10 = 8
\]

Isolate \( x \) by subtracting 10 from both sides:

\[
-0.25x = 8 - 10
\]
\[
-0.25x = -2
\]

Solve for \( x \) by dividing both sides by -0.25:

\[
x = \frac{-2}{-0.25} = 8
\]

Thus, \( x = 8 \) mL of the 25% acid solution. The amount of the 50% acid solution is then:

\[
20 - x = 20 - 8 = 12 \text{ mL}
\]

To verify, calculate the total acid content:

\[
0.25 \times 8 + 0.50 \times 12 = 2 + 6 = 8 \text{ mL}
\]

The final solution is 20 mL with 8 mL of acid, which is 40% acid.

Therefore, the solution requires:

\[
\boxed{8 \text{ mL of 25\% solution and } 12 \text{ mL of 50\% solution}}
\]

System: You are a helpful assistant.


[User] How many ways are there to distribute 28 identical apples to 4 people such that each person receives at least one apple?

[Assistant]
To determine the number of ways to distribute 28 identical apples to 4 people such that each person receives at least one apple, we can use the stars and bars method with an adjustment for the constraint that each person gets at least one apple.

1. **Adjust for the constraint**: Since each person must receive at least one apple, we first give each person one apple. This uses up \(4\) apples, leaving us with \(28 - 4 = 24\) apples to distribute freely among the 4 people.

2. **Formulate the problem**: The problem now reduces to finding the number of non-negative integer solutions to the equation \(x_1 + x_2 + x_3 + x_4 = 24\), where \(x_i\) represents the number of additional apples the \(i\)-th person receives.

3. **Apply the stars and bars formula**: The number of non-negative integer solutions to the equation \(x_1 + x_2 + x_3 + x_4 = 24\) is given by the combination formula \(\binom{n + k - 1}{k - 1}\), where \(n = 24\) (remaining apples) and \(k = 4\) (people).

\[
\binom{24 + 4 - 1}{4 - 1} = \binom{27}{3}
\]

4. **Calculate the combination**:

\[
\binom{27}{3} = \frac{27 \times 26 \times 25}{3 \times 2 \times 1} = \frac{17550}{6} = 2925
\]

Thus, the number of ways to distribute the 28 apples to 4 people such that each person receives at least one apple is \(\boxed{2925}\).

System: You are a helpful assistant.


[User] Write down a definite integral that will give the value of the area under the curve \( y = x^2 \cos\left(\frac{1}{5}x\right) \) between \( x = \frac{1}{2}\pi \) and \( x = \frac{5}{2}\pi \). (The expression \( x^2 \cos\left(\frac{1}{5}x\right) \) takes no negative values for \( \frac{1}{2}\pi \leq x \leq \frac{5}{2}\pi \). You are not asked to evaluate the integral by hand.)

[Assistant]
To find the definite integral that gives the area under the curve \( y = x^2 \cos\left(\frac{1}{5}x\right) \) between \( x = \frac{1}{2}\pi \) and \( x = \frac{5}{2}\pi \), we follow these steps:

1. **Identify the function and the interval**: The function given is \( y = x^2 \cos\left(\frac{1}{5}x\right) \), and the interval is from \( x = \frac{1}{2}\pi \) to \( x = \frac{5}{2}\pi \).

2. **Verify the non-negativity of the function**: The problem states that \( x^2 \cos\left(\frac{1}{5}x\right) \) does not take any negative values in the interval \( \frac{1}{2}\pi \leq x \leq \frac{5}{2}\pi \). To confirm this, we note that \( x^2 \) is always non-negative, and \( \cos\left(\frac{1}{5}x\right) \) is non-negative when \( \frac{1}{5}x \) lies between \( \frac{\pi}{10} \) and \( \frac{\pi}{2} \). Since \( \frac{1}{5}x \) ranges from \( \frac{\pi}{10} \) to \( \frac{\pi}{2} \) within the given interval, \( \cos\left(\frac{1}{5}x\right) \) is indeed non-negative.

3. **Set up the definite integral**: The area under the curve \( y = x^2 \cos\left(\frac{1}{5}x\right) \) from \( x = \frac{1}{2}\pi \) to \( x = \frac{5}{2}\pi \) is given by the definite integral:
   \[
   \int_{\pi/2}^{5\pi/2} x^2 \cos\left(\frac{x}{5}\right) \, dx
   \]

Thus, the definite integral that gives the area under the curve \( y = x^2 \cos\left(\frac{1}{5}x\right) \) between \( x = \frac{1}{2}\pi \) and \( x = \frac{5}{2}\pi \) is:
\[
\boxed{\int_{\pi/2}^{5\pi/2} x^2 \cos\left(\frac{x}{5}\right) \, dx}
\]

System: You are a helpful assistant.


[User] An investment of $\$$10,000 is made in a government bond that will pay 6$\%$ interest compounded annually. At the end of five years, what is the total number of dollars in this investment? Express your answer to the nearest whole number.

[Assistant]
To determine the total amount of an investment of \$10,000 after 5 years with 6% annual compound interest, we use the compound interest formula:

\[ A = P (1 + r)^t \]

where:
- \( P \) is the principal amount (\$10,000),
- \( r \) is the annual interest rate (6%, or 0.06),
- \( t \) is the number of years (5).

First, we calculate the multiplier:

\[ (1 + r)^t = (1 + 0.06)^5 = 1.06^5 \]

Next, we compute \( 1.06^5 \):

\[ 1.06^1 = 1.06 \]
\[ 1.06^2 = 1.06 \times 1.06 = 1.1236 \]
\[ 1.06^3 = 1.1236 \times 1.06 = 1.191016 \]
\[ 1.06^4 = 1.191016 \times 1.06 = 1.26247696 \]
\[ 1.06^5 = 1.26247696 \times 1.06 = 1.3382255776 \]

Now, we multiply the principal by this multiplier:

\[ A = 10,000 \times 1.3382255776 = 13,382.255776 \]

Finally, we round the result to the nearest whole number:

\[ 13,382.255776 \approx 13,382 \]

Therefore, the total amount after 5 years is:

\[
\boxed{13382}
\]

System: You are a helpful assistant.


[User] Morgan walks \( x \) yards north, then walks \( (2x-1) \) yards west. She is now \( (2x+1) \) yards from her starting point. What is the distance, in yards, she travels?

[Assistant]
To solve the problem, we start by noting that Morgan's path forms a right triangle. The northward distance is \( x \) yards, the westward distance is \( (2x - 1) \) yards, and the straight-line distance from her starting point is \( (2x + 1) \) yards. Using the Pythagorean theorem, we can set up the following equation:

\[
x^2 + (2x - 1)^2 = (2x + 1)^2
\]

First, we expand the terms:

\[
(2x - 1)^2 = 4x^2 - 4x + 1
\]
\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]

Substituting these into the equation, we get:

\[
x^2 + 4x^2 - 4x + 1 = 4x^2 + 4x + 1
\]

Combining like terms on the left side:

\[
5x^2 - 4x + 1 = 4x^2 + 4x + 1
\]

Next, we move all terms to one side of the equation:

\[
5x^2 - 4x + 1 - 4x^2 - 4x - 1 = 0
\]

Simplifying:

\[
x^2 - 8x = 0
\]

Factoring out \( x \):

\[
x(x - 8) = 0
\]

This gives us two potential solutions: \( x = 0 \) or \( x = 8 \). Since \( x = 0 \) does not make sense in the context of the problem (as it would imply no movement), we have:

\[
x = 8
\]

Now, we calculate the total distance Morgan traveled. The northward distance is \( x = 8 \) yards, and the westward distance is \( 2x - 1 = 2(8) - 1 = 15 \) yards. Therefore, the total distance traveled is:

\[
x + (2x - 1) = 8 + 15 = 23 \text{ yards}
\]

Thus, the distance Morgan travels is:

\[
\boxed{23}
\]

System: You are a helpful assistant.


[User] Given a number with last digit \( a \) in base \( b \), consider the sequence \( 1, a, a^2, a^3, \ldots \). What is the period of the last digits of this sequence for \( a = 17 \) and \( b = 10 \)?

[Assistant]
To find the period of the last digits of the sequence \(1, 17, 17^2, 17^3, \ldots\) in base 10, we need to consider the last digit of each term in the sequence. The last digit of 17 is 7, so we are essentially looking at the sequence \(1, 7, 7^2, 7^3, \ldots\) and determining the period of the last digits of these powers of 7 modulo 10.

Let's compute the last digits of the first few powers of 7:

\[
\begin{aligned}
7^1 & = 7 \quad \text{(last digit is 7)} \\
7^2 & = 49 \quad \text{(last digit is 9)} \\
7^3 & = 343 \quad \text{(last digit is 3)} \\
7^4 & = 2401 \quad \text{(last digit is 1)} \\
7^5 & = 16807 \quad \text{(last digit is 7)} \\
\end{aligned}
\]

We observe that the last digits repeat every 4 terms: \(7, 9, 3, 1\). Therefore, the period of the last digits of the sequence \(1, 7, 7^2, 7^3, \ldots\) is 4.

Thus, the period of the last digits of the sequence \(1, 17, 17^2, 17^3, \ldots\) in base 10 is \(\boxed{4}\).

System: You are a helpful assistant.


[User] The area of a rectangle is 432 square centimeters. What would the new area be if the length of the rectangle was decreased by $10\%$ and the width of the rectangle was increased by $10\%$? Express your answer to the nearest whole number.

[Assistant]
To solve the problem, we start by noting that the area of the original rectangle is given by:

\[ A = L \times W = 432 \text{ square centimeters} \]

where \( L \) is the length and \( W \) is the width of the rectangle.

According to the problem, the length is decreased by 10%, and the width is increased by 10%. Therefore, the new length and new width are:

\[ \text{New length} = 0.9L \]
\[ \text{New width} = 1.1W \]

The new area of the rectangle can be calculated by multiplying the new length and the new width:

\[ \text{New area} = (0.9L) \times (1.1W) \]

We can simplify this expression:

\[ \text{New area} = 0.9 \times 1.1 \times L \times W \]

Since \( L \times W = 432 \):

\[ \text{New area} = 0.9 \times 1.1 \times 432 \]

Next, we calculate the product of 0.9 and 1.1:

\[ 0.9 \times 1.1 = 0.99 \]

So the new area is:

\[ \text{New area} = 0.99 \times 432 \]

Now, we perform the multiplication:

\[ 0.99 \times 432 = 427.68 \]

Finally, we round 427.68 to the nearest whole number:

\[ 427.68 \approx 428 \]

Therefore, the new area of the rectangle, rounded to the nearest whole number, is:

\[
\boxed{428}
\]

System: You are a helpful assistant.


[User] Calculate \( \frac{d}{d \epsilon } \left( \| \dot{ \mathbf{x} } ^{ \epsilon } \| ^{2} \right) \) where \( \dot{ \mathbf{x} } ^{ \epsilon } = \frac{ \mathbf{x}(t) + \epsilon \mathbf{z} (t) }{ \| \mathbf{x} (t) + \epsilon \mathbf{z} (t) \| } \) at the point where \( \epsilon = 0 \).

[Assistant]
To calculate the derivative \( \frac{d}{d \epsilon} \left( \| \dot{\mathbf{x}}^\epsilon \|^2 \right) \) at \( \epsilon = 0 \), we start with the given vector \( \dot{\mathbf{x}}^\epsilon \):

\[
\dot{\mathbf{x}}^\epsilon = \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|}
\]

The norm squared of \( \dot{\mathbf{x}}^\epsilon \) is:

\[
\| \dot{\mathbf{x}}^\epsilon \|^2 = \left( \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|} \right) \cdot \left( \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|} \right)
\]

Simplifying the expression, we get:

\[
\| \dot{\mathbf{x}}^\epsilon \|^2 = \frac{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|^2}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|^2} = 1
\]

Since the norm squared of \( \dot{\mathbf{x}}^\epsilon \) is always 1 for all \( \epsilon \) (as long as \( \mathbf{x}(t) + \epsilon \mathbf{z}(t) \neq \mathbf{0} \)), the function \( \| \dot{\mathbf{x}}^\epsilon \|^2 \) is a constant function. The derivative of a constant function with respect to \( \epsilon \) is zero. Therefore, evaluating the derivative at \( \epsilon = 0 \) gives:

\[
\frac{d}{d \epsilon} \left( \| \dot{\mathbf{x}}^\epsilon \|^2 \right) \bigg|_{\epsilon = 0} = 0
\]

Thus, the final answer is:

\[
\boxed{0}
\]

System: You are a helpful assistant.


[User] How many triangles can be formed by connecting 5 points on one line and 12 points on a parallel line?

[Assistant]
To determine the number of triangles that can be formed by connecting 5 points on one line and 12 points on a parallel line, we need to ensure that the three points chosen are not collinear. Since the points are on two parallel lines, a triangle can be formed if and only if two points are chosen from one line and one point is chosen from the other line.

Let's denote the number of points on the first line as \( A = 5 \) and the number of points on the second line as \( B = 12 \).

1. **Choosing 2 points from the first line and 1 point from the second line:**
   - The number of ways to choose 2 points from 5 points on the first line is given by the combination formula \( \binom{5}{2} \).
   - The number of ways to choose 1 point from 12 points on the second line is given by \( \binom{12}{1} \).

   Therefore, the total number of ways to form triangles in this configuration is:
   \[
   \binom{5}{2} \times \binom{12}{1} = \frac{5!}{2!(5-2)!} \times 12 = \frac{5 \times 4}{2 \times 1} \times 12 = 10 \times 12 = 120
   \]

2. **Choosing 2 points from the second line and 1 point from the first line:**
   - The number of ways to choose 2 points from 12 points on the second line is given by \( \binom{12}{2} \).
   - The number of ways to choose 1 point from 5 points on the first line is given by \( \binom{5}{1} \).

   Therefore, the total number of ways to form triangles in this configuration is:
   \[
   \binom{12}{2} \times \binom{5}{1} = \frac{12!}{2!(12-2)!} \times 5 = \frac{12 \times 11}{2 \times 1} \times 5 = 66 \times 5 = 330
   \]

3. **Summing the two cases:**
   - The total number of triangles is the sum of the two configurations:
   \[
   120 + 330 = 450
   \]

Thus, the total number of triangles that can be formed is \(\boxed{450}\).

System: You are a helpful assistant.


[User] Given that \( m \) and \( n \) are positive integers and \( m^2 + 5n^2 = 36 \), find the value of \( m + 5n \).

[Assistant]
To solve the equation \( m^2 + 5n^2 = 36 \) for positive integers \( m \) and \( n \), we will systematically test possible values for \( n \) and determine if the corresponding \( m \) is an integer.

1. **Testing \( n = 1 \):**
   \[
   5n^2 = 5(1)^2 = 5
   \]
   \[
   m^2 = 36 - 5 = 31
   \]
   Since 31 is not a perfect square, \( m \) is not an integer. Thus, \( n = 1 \) is not a solution.

2. **Testing \( n = 2 \):**
   \[
   5n^2 = 5(2)^2 = 20
   \]
   \[
   m^2 = 36 - 20 = 16
   \]
   Since \( 16 \) is a perfect square, \( m = 4 \). Thus, \( n = 2 \) and \( m = 4 \) is a solution.

3. **Testing \( n \geq 3 \):**
   \[
   5n^2 \geq 5(3)^2 = 45
   \]
   Since \( 5n^2 \) exceeds 36 for \( n \geq 3 \), \( m^2 \) would be negative, which is not possible for positive integers.

Thus, the only valid solution is \( m = 4 \) and \( n = 2 \).

Finally, we calculate \( m + 5n \):
\[
m + 5n = 4 + 5(2) = 4 + 10 = 14
\]

Therefore, the value of \( m + 5n \) is \(\boxed{14}\).

System: You are a helpful assistant.


[User] Given $\log 1.23 = 0.07$, find $10^{1.07}$.

[Assistant]
To find \(10^{1.07}\) given that \(\log 1.23 = 0.07\), we can proceed as follows:

1. Recognize that \(\log 1.23 = 0.07\) implies \(\log_{10} 1.23 = 0.07\). By the definition of logarithms, this means:
   \[
   10^{0.07} = 1.23
   \]

2. We need to find \(10^{1.07}\). We can express \(1.07\) as \(1 + 0.07\). Therefore:
   \[
   10^{1.07} = 10^{1 + 0.07}
   \]

3. Using the properties of exponents, we can split the exponent:
   \[
   10^{1 + 0.07} = 10^1 \cdot 10^{0.07}
   \]

4. We know from the given information that \(10^{0.07} = 1.23\). Also, \(10^1 = 10\). Therefore:
   \[
   10^{1.07} = 10 \cdot 1.23
   \]

5. Perform the multiplication:
   \[
   10 \cdot 1.23 = 12.3
   \]

Thus, the value of \(10^{1.07}\) is \(\boxed{12.3}\).

System: You are a helpful assistant.


[User] Given parallelogram ABCD, with AB = 10, AD = 12, and BD = $2\sqrt{13}$, find AC.

[Assistant]
To find the length of \( AC \) in parallelogram \( ABCD \) where \( AB = 10 \), \( AD = 12 \), and \( BD = 2\sqrt{13} \), we can use the property of parallelograms that states the sum of the squares of the diagonals is equal to the sum of the squares of all the sides. The formula is:

\[
AC^2 + BD^2 = 2(AB^2 + AD^2)
\]

First, we compute \( BD^2 \):

\[
BD = 2\sqrt{13} \implies BD^2 = (2\sqrt{13})^2 = 4 \times 13 = 52
\]

Next, we compute \( AB^2 \) and \( AD^2 \):

\[
AB = 10 \implies AB^2 = 10^2 = 100
\]
\[
AD = 12 \implies AD^2 = 12^2 = 144
\]

Now, we substitute these values into the formula:

\[
AC^2 + 52 = 2(100 + 144)
\]

Calculate the sum inside the parentheses:

\[
100 + 144 = 244
\]

Then multiply by 2:

\[
2 \times 244 = 488
\]

So the equation becomes:

\[
AC^2 + 52 = 488
\]

Subtract 52 from both sides to isolate \( AC^2 \):

\[
AC^2 = 488 - 52 = 436
\]

Finally, take the square root of both sides to find \( AC \):

\[
AC = \sqrt{436}
\]

Factor 436 to simplify the square root:

\[
436 = 4 \times 109 \implies \sqrt{436} = \sqrt{4 \times 109} = 2\sqrt{109}
\]

Therefore, the length of \( AC \) is:

\[
\boxed{2\sqrt{109}}
\]

System: You are a helpful assistant.


[User] A pirate is counting the loot he has stolen from S.S. AOPS, on which everything is counted in base 5.  A treasure chest contains $3124_{5}$ dollars worth of jewelry, $3122_{5}$ dollars worth of gold coins, and $124_{5}$ dollars worth of rubbing alcohol.  What is the total dollar amount that this pirate has accumulated? Express your answer in base 10.

[Assistant]
To find the total dollar amount the pirate has accumulated, we need to convert each base 5 value to base 10 and then sum the results.

1. **Convert Jewelry (3124₅) to base 10:**
   \[
   3 \times 5^3 + 1 \times 5^2 + 2 \times 5^1 + 4 \times 5^0 = 3 \times 125 + 1 \times 25 + 2 \times 5 + 4 \times 1 = 375 + 25 + 10 + 4 = 414_{10}
   \]

2. **Convert Gold Coins (3122₅) to base 10:**
   \[
   3 \times 5^3 + 1 \times 5^2 + 2 \times 5^1 + 2 \times 5^0 = 3 \times 125 + 1 \times 25 + 2 \times 5 + 2 \times 1 = 375 + 25 + 10 + 2 = 412_{10}
   \]

3. **Convert Rubbing Alcohol (124₅) to base 10:**
   \[
   1 \times 5^2 + 2 \times 5^1 + 4 \times 5^0 = 1 \times 25 + 2 \times 5 + 4 \times 1 = 25 + 10 + 4 = 39_{10}
   \]

Next, we sum the converted values in base 10:
\[
414 + 412 + 39 = 865
\]

Thus, the total dollar amount the pirate has accumulated is \(\boxed{865}\).

System: You are a helpful assistant.


[User] What value of $k$ will make $x^2 - 16x + k$ the square of a binomial?

[Assistant]
To determine the value of \( k \) that makes the quadratic expression \( x^2 - 16x + k \) the square of a binomial, we need to recognize the form of a perfect square trinomial. A perfect square trinomial can be written as \( (x - a)^2 \), which expands to \( x^2 - 2ax + a^2 \).

Given the expression \( x^2 - 16x + k \), we can compare it to the expanded form \( x^2 - 2ax + a^2 \). By comparing the coefficients of the \( x \)-terms, we see that:
\[ -2a = -16. \]
Solving for \( a \), we get:
\[ a = \frac{-16}{-2} = 8. \]

Next, we need to find the constant term \( k \), which is \( a^2 \). Since \( a = 8 \), we have:
\[ k = a^2 = 8^2 = 64. \]

To verify, we can substitute \( k = 64 \) back into the original expression and check if it forms a perfect square:
\[ x^2 - 16x + 64 = (x - 8)^2. \]
This confirms that the expression is indeed a perfect square when \( k = 64 \).

Thus, the value of \( k \) that makes \( x^2 - 16x + k \) the square of a binomial is \(\boxed{64}\).

System: You are a helpful assistant.


[User] If the sides of a cyclic quadrilateral are the roots of the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), express the sum of the sides \(S\) in terms of \(p\), \(q\), \(s\), and \(t\).

[Assistant]
To find the sum of the sides \(S\) of a cyclic quadrilateral whose sides are the roots of the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), we can use Vieta's formulas.

Given the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), let the roots be \(a\), \(b\), \(c\), and \(d\). According to Vieta's formulas, the sum of the roots of the polynomial is given by the coefficient of the \(x^3\) term with a sign change. Specifically, for the polynomial \(x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0\), the sum of the roots is \(-a_3\).

In our case, the polynomial is \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), so the coefficient of \(x^3\) is \(-2s\). Therefore, the sum of the roots \(a + b + c + d\) is:

\[
a + b + c + d = -(-2s) = 2s
\]

Thus, the sum of the sides \(S\) of the cyclic quadrilateral is:

\[
\boxed{2s}
\]

System: You are a helpful assistant.


[User] Solve the equation $10^{2y} = 25$ using only laws of exponents and algebra.

[Assistant]
To solve the equation \(10^{2y} = 25\) using laws of exponents and algebra, follow these steps:

1. Recognize that \(25\) can be written as \(5^2\). Thus, the equation becomes:
   \[
   10^{2y} = 5^2
   \]

2. Take the logarithm (base 10) of both sides of the equation:
   \[
   \log(10^{2y}) = \log(5^2)
   \]

3. Apply the logarithm power rule, which states \(\log(a^b) = b \log(a)\):
   \[
   2y \cdot \log(10) = 2 \log(5)
   \]

4. Since \(\log(10) = 1\), the equation simplifies to:
   \[
   2y = 2 \log(5)
   \]

5. Divide both sides by 2 to solve for \(y\):
   \[
   y = \log(5)
   \]

6. Verify the solution by substituting \(y = \log(5)\) back into the original equation:
   \[
   10^{2 \log(5)} = 10^{\log(5^2)} = 10^{\log(25)} = 25
   \]

Thus, the solution to the equation \(10^{2y} = 25\) is:
\[
\boxed{\log 5}
\]

System: You are a helpful assistant.


[User] Find the general solution of the differential equation $ty''(t) + 2y'(t) = 0$ using the substitution $y'(t) = u(t)$.

[Assistant]
To solve the differential equation \( t y''(t) + 2 y'(t) = 0 \) using the substitution \( y'(t) = u(t) \), we proceed as follows:

1. **Substitution:**
   Let \( y'(t) = u(t) \). Then \( y''(t) = u'(t) \).

2. **Rewrite the Equation:**
   Substitute \( y'(t) \) and \( y''(t) \) into the original equation:
   \[
   t u'(t) + 2 u(t) = 0
   \]

3. **Solve the First-Order Linear ODE:**
   This is a first-order linear ordinary differential equation. We can solve it by separating variables or using an integrating factor. Here, we use the integrating factor method.

   The standard form of a first-order linear ODE is \( u'(t) + P(t) u(t) = Q(t) \). For our equation, \( P(t) = \frac{2}{t} \) and \( Q(t) = 0 \).

   The integrating factor \( \mu(t) \) is given by:
   \[
   \mu(t) = e^{\int \frac{2}{t} \, dt} = e^{2 \ln t} = t^2
   \]

4. **Multiply by the Integrating Factor:**
   Multiply both sides of the equation \( t u'(t) + 2 u(t) = 0 \) by \( t^2 \):
   \[
   t^2 u'(t) + 2 t u(t) = 0
   \]

   Notice that the left-hand side is the derivative of \( t^2 u(t) \):
   \[
   \frac{d}{dt} (t^2 u(t)) = 0
   \]

5. **Integrate:**
   Integrate both sides with respect to \( t \):
   \[
   t^2 u(t) = C_1
   \]
   where \( C_1 \) is the constant of integration. Therefore,
   \[
   u(t) = \frac{C_1}{t^2}
   \]

6. **Integrate to Find \( y(t) \):**
   Recall that \( u(t) = y'(t) \). Integrate \( u(t) \) to find \( y(t) \):
   \[
   y(t) = \int \frac{C_1}{t^2} \, dt = -\frac{C_1}{t} + C_2
   \]
   where \( C_2 \) is another constant of integration.

7. **General Solution:**
   The general solution to the differential equation is:
   \[
   y(t) = -\frac{C_1}{t} + C_2
   \]

   To match the standard form, we can write:
   \[
   y(t) = \frac{A}{t} + B
   \]
   where \( A = -C_1 \) and \( B = C_2 \).

Thus, the general solution is:
\[
\boxed{y(t) = \frac{A}{t} + B}
\]
where \( A \) and \( B \) are constants of integration.

System: You are a helpful assistant.


[User] How many multiples of 10 are between 9 and 101?

[Assistant]
To determine how many multiples of 10 are between 9 and 101, we first identify the smallest and largest multiples of 10 within this range.

The smallest multiple of 10 greater than 9 is 10.
The largest multiple of 10 less than or equal to 101 is 100.

The multiples of 10 in this range form an arithmetic sequence: 10, 20, 30, ..., 100.

The formula for the number of terms in an arithmetic sequence is given by:
\[
\text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1
\]

Here, the first term (\(a\)) is 10, the last term (\(l\)) is 100, and the common difference (\(d\)) is 10.

Plugging in these values:
\[
\text{Number of terms} = \frac{100 - 10}{10} + 1 = \frac{90}{10} + 1 = 9 + 1 = 10
\]

Thus, there are 10 multiples of 10 between 9 and 101.

\[
\boxed{10}
\]

System: You are a helpful assistant.


[User] If $\sqrt{5 + x} + \sqrt{20 - x} = 7$, what is the value of $(5 + x)(20 - x)$?

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

\[
\sqrt{5 + x} + \sqrt{20 - x} = 7
\]

Let \( A = \sqrt{5 + x} \) and \( B = \sqrt{20 - x} \). Therefore, the equation becomes:

\[
A + B = 7
\]

We are asked to find the value of \( (5 + x)(20 - x) \). Notice that:

\[
(5 + x) = A^2 \quad \text{and} \quad (20 - x) = B^2
\]

Thus, we need to find \( A^2 B^2 \), which is \( (AB)^2 \).

First, square both sides of the equation \( A + B = 7 \):

\[
(A + B)^2 = 7^2
\]

This expands to:

\[
A^2 + 2AB + B^2 = 49
\]

We know that:

\[
A^2 + B^2 = (5 + x) + (20 - x) = 25
\]

Substitute \( A^2 + B^2 = 25 \) into the expanded equation:

\[
25 + 2AB = 49
\]

Solve for \( 2AB \):

\[
2AB = 49 - 25 = 24
\]

Thus,

\[
AB = \frac{24}{2} = 12
\]

Now, we need to find \( (AB)^2 \):

\[
(AB)^2 = 12^2 = 144
\]

Therefore, the value of \( (5 + x)(20 - x) \) is:

\[
\boxed{144}
\]

System: You are a helpful assistant.


[User] Solve for \( x \) in the congruence equation \( 100 + 7x \equiv 9 \pmod{11} \).

[Assistant]
To solve the congruence equation \( 100 + 7x \equiv 9 \pmod{11} \), follow these steps:

1. **Simplify the equation**:
   \[
   100 + 7x \equiv 9 \pmod{11}
   \]
   Subtract 100 from both sides:
   \[
   7x \equiv 9 - 100 \pmod{11}
   \]
   Simplify \(9 - 100\):
   \[
   9 - 100 = -91
   \]
   Therefore:
   \[
   7x \equiv -91 \pmod{11}
   \]

2. **Simplify \(-91 \mod 11\)**:
   \[
   -91 \equiv 8 \pmod{11}
   \]
   (Since \(91 = 8 \times 11 + 3\), thus \(-91 = -8 \times 11 + 8\), so \(-91 \equiv 8 \pmod{11}\)).

   So the equation becomes:
   \[
   7x \equiv 8 \pmod{11}
   \]

3. **Find the multiplicative inverse of 7 modulo 11**:
   The multiplicative inverse of 7 modulo 11 is a number \(y\) such that:
   \[
   7y \equiv 1 \pmod{11}
   \]
   By trial:
   \[
   7 \times 1 = 7 \pmod{11} \quad (\text{not } 1)
   \]
   \[
   7 \times 2 = 14 \equiv 3 \pmod{11} \quad (\text{not } 1)
   \]
   \[
   7 \times 3 = 21 \equiv 10 \pmod{11} \quad (\text{not } 1)
   \]
   \[
   7 \times 4 = 28 \equiv 6 \pmod{11} \quad (\text{not } 1)
   \]
   \[
   7 \times 5 = 35 \equiv 2 \pmod{11} \quad (\text{not } 1)
   \]
   \[
   7 \times 8 = 56 \equiv 1 \pmod{11} \quad (\text{correct})
   \]
   Therefore, the inverse of 7 modulo 11 is 8.

4. **Multiply both sides of the equation by the inverse**:
   \[
   7x \equiv 8 \pmod{11}
   \]
   Multiply both sides by 8:
   \[
   x \equiv 8 \times 8 \pmod{11}
   \]
   Simplify \(8 \times 8\):
   \[
   8 \times 8 = 64
   \]
   \[
   64 \mod 11 = 64 - 5 \times 11 = 64 - 55 = 9
   \]
   Therefore:
   \[
   x \equiv 9 \pmod{11}
   \]

The solution to the congruence equation is:
\[
\boxed{9}
\]

System: You are a helpful assistant.


[User] How many pages are there in a book if it takes 852 digits to number the pages consecutively?

[Assistant]
To determine the number of pages in a book given that it takes 852 digits to number the pages consecutively, we can break down the problem into different ranges of page numbers and calculate the number of digits used in each range.

1. **Pages 1 to 9:**
   - There are 9 pages.
   - Each page uses 1 digit.
   - Total digits used: \( 9 \times 1 = 9 \) digits.

2. **Pages 10 to 99:**
   - There are \( 99 - 10 + 1 = 90 \) pages.
   - Each page uses 2 digits.
   - Total digits used: \( 90 \times 2 = 180 \) digits.

3. **Total digits used for pages 1 to 99:**
   - Sum of digits from 1 to 9 and 10 to 99: \( 9 + 180 = 189 \) digits.

4. **Remaining digits:**
   - Total digits given: 852.
   - Digits used for pages 1 to 99: 189.
   - Remaining digits: \( 852 - 189 = 663 \) digits.

5. **Pages 100 onwards:**
   - Each page uses 3 digits.
   - Number of pages: \( \frac{663}{3} = 221 \) pages.

6. **Total number of pages:**
   - Pages from 1 to 99: 99 pages.
   - Pages from 100 onwards: 221 pages.
   - Total pages: \( 99 + 221 = 320 \) pages.

Thus, the number of pages in the book is \(\boxed{320}\).

System: You are a helpful assistant.


[User] Evaluate the limit $\lim _{ n\rightarrow \infty  }{ \frac { { 1 }^{ 99 }+{ 2 }^{ 99 }+{ 3 }^{ 99 }+{ 4 }^{ 99 }+.................+{ n }^{ 99 } }{ { n }^{ 100 } }  }$.

[Assistant]
To evaluate the limit \(\lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \cdots + n^{99}}{n^{100}}\), we can use the concept of Riemann sums or the asymptotic behavior of polynomial sums.

### Approach Using Riemann Sums

1. **Rewrite the Sum**:
   Consider the sum \(1^{99} + 2^{99} + \cdots + n^{99}\). We can rewrite it as:
   \[
   \sum_{k=1}^n k^{99}
   \]

2. **Factor Out \(n^{99}\)**:
   Factor out \(n^{99}\) from each term in the sum:
   \[
   \sum_{k=1}^n k^{99} = n^{99} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}
   \]

3. **Form the Limit**:
   Now, divide by \(n^{100}\):
   \[
   \frac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}} = \frac{n^{99} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}}{n^{100}} = \frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}
   \]

4. **Recognize the Riemann Sum**:
   The expression \(\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}\) is a Riemann sum for the function \(f(x) = x^{99}\) over the interval \([0, 1]\).

5. **Evaluate the Integral**:
   As \(n \to \infty\), the Riemann sum converges to the integral:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99} = \int_0^1 x^{99} \, dx
   \]

6. **Compute the Integral**:
   Evaluate the integral:
   \[
   \int_0^1 x^{99} \, dx = \left[ \frac{x^{100}}{100} \right]_0^1 = \frac{1^{100}}{100} - \frac{0^{100}}{100} = \frac{1}{100}
   \]

### Conclusion

Thus, the limit is:
\[
\lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \cdots + n^{99}}{n^{100}} = \boxed{\frac{1}{100}}
\]

System: You are a helpful assistant.


[User] Given three complex numbers \( z_1, z_2, \) and \( z_3 \) such that \( |z_1| = |z_2| = |z_3| = 1 \) and \( \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1 \), find \( |z_1 + z_2 + z_3| \).

[Assistant]
Given three complex numbers \( z_1, z_2, \) and \( z_3 \) such that \( |z_1| = |z_2| = |z_3| = 1 \) and \( \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1 \), we need to find \( |z_1 + z_2 + z_3| \).

Since \( |z_1| = |z_2| = |z_3| = 1 \), each \( z_i \) lies on the unit circle. For any complex number \( z \) on the unit circle, the reciprocal \( \frac{1}{z} \) is the complex conjugate of \( z \), denoted \( \overline{z} \). Therefore, we can rewrite the given condition as:
\[
\left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = \left| \overline{z_1} + \overline{z_2} + \overline{z_3} \right|.
\]

The absolute value of a complex number and its conjugate are equal, so:
\[
\left| \overline{z_1} + \overline{z_2} + \overline{z_3} \right| = \left| z_1 + z_2 + z_3 \right|.
\]

Given that \( \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1 \), it follows that:
\[
\left| z_1 + z_2 + z_3 \right| = 1.
\]

Thus, the value of \( |z_1 + z_2 + z_3| \) is \(\boxed{1}\).

System: You are a helpful assistant.


[User] Suppose we have a Gaussian random variable with mean = 1 and variance = 1. We create \( N \) samples \( \chi_1, \chi_2, \ldots, \chi_N \) from this distribution and calculate the arithmetic average \( \mu = \frac{\chi_1 + \chi_2 + \ldots + \chi_N}{N} \). What is the mean and variance of \( \mu \)?

[Assistant]
To find the mean and variance of the arithmetic average \(\mu\) of \(N\) samples \(\chi_1, \chi_2, \ldots, \chi_N\) from a Gaussian distribution with mean 1 and variance 1, we proceed as follows:

1. **Mean Calculation:**
   The arithmetic average \(\mu\) is given by:
   \[
   \mu = \frac{\chi_1 + \chi_2 + \ldots + \chi_N}{N}
   \]
   The expected value (mean) of \(\mu\) is:
   \[
   E[\mu] = E\left[\frac{\chi_1 + \chi_2 + \ldots + \chi_N}{N}\right]
   \]
   Using the linearity of expectation, we have:
   \[
   E[\mu] = \frac{E[\chi_1] + E[\chi_2] + \ldots + E[\chi_N]}{N}
   \]
   Since each \(\chi_i\) has a mean of 1, we get:
   \[
   E[\mu] = \frac{1 + 1 + \ldots + 1}{N} = \frac{N \cdot 1}{N} = 1
   \]
   Therefore, the mean of \(\mu\) is:
   \[
   \text{Mean} = 1
   \]

2. **Variance Calculation:**
   The variance of \(\mu\) is given by:
   \[
   \text{Var}(\mu) = \text{Var}\left(\frac{\chi_1 + \chi_2 + \ldots + \chi_N}{N}\right)
   \]
   Using the property of variance for independent random variables, we have:
   \[
   \text{Var}(\mu) = \text{Var}\left(\frac{1}{N} (\chi_1 + \chi_2 + \ldots + \chi_N)\right)
   \]
   The variance of a sum of independent random variables is the sum of their variances, and scaling by a constant factor \( \frac{1}{N} \) results in scaling the variance by \( \left(\frac{1}{N}\right)^2 \):
   \[
   \text{Var}(\mu) = \left(\frac{1}{N}\right)^2 \text{Var}(\chi_1 + \chi_2 + \ldots + \chi_N)
   \]
   Since each \(\chi_i\) has a variance of 1, we get:
   \[
   \text{Var}(\chi_1 + \chi_2 + \ldots + \chi_N) = \text{Var}(\chi_1) + \text{Var}(\chi_2) + \ldots + \text{Var}(\chi_N) = N \cdot 1 = N
   \]
   Therefore:
   \[
   \text{Var}(\mu) = \left(\frac{1}{N}\right)^2 \cdot N = \frac{N}{N^2} = \frac{1}{N}
   \]
   Thus, the variance of \(\mu\) is:
   \[
   \text{Variance} = \frac{1}{N}
   \]

The final answer is:
\[
\boxed{\text{Mean} = 1, \ \text{Variance} = \frac{1}{N}}
\]

System: You are a helpful assistant.


[User] What is the expected value of the roll of a standard 6-sided die?

[Assistant]
To find the expected value of rolling a standard 6-sided die, we need to consider each possible outcome and its associated probability. A standard 6-sided die has outcomes \(1, 2, 3, 4, 5,\) and \(6\), each with an equal probability of \(\frac{1}{6}\).

The expected value \(E\) is calculated by summing the products of each outcome and its probability:

\[
E = \sum_{i=1}^{6} i \cdot \frac{1}{6}
\]

Breaking this down:

\[
E = \left(1 \cdot \frac{1}{6}\right) + \left(2 \cdot \frac{1}{6}\right) + \left(3 \cdot \frac{1}{6}\right) + \left(4 \cdot \frac{1}{6}\right) + \left(5 \cdot \frac{1}{6}\right) + \left(6 \cdot \frac{1}{6}\right)
\]

Simplify each term:

\[
E = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6}
\]

Combine the fractions:

\[
E = \frac{1 + 2 + 3 + 4 + 5 + 6}{6}
\]

Calculate the sum of the numerators:

\[
1 + 2 + 3 + 4 + 5 + 6 = 21
\]

Thus:

\[
E = \frac{21}{6} = 3.5
\]

Therefore, the expected value of the roll of a standard 6-sided die is \(\boxed{3.5}\).

System: You are a helpful assistant.