[User] If \((1 + x + x^2)^n = a_0 + a_1 x + \cdots + a_{2n} x^{2n}\), find \(a_0 + a_2 + a_4 + \cdots + a_{2n}\).

[Assistant]
To find the sum of the even-indexed coefficients in the expansion of \((1 + x + x^2)^n\), we start by evaluating the polynomial at \(x = 1\) and \(x = -1\).

1. Substituting \(x = 1\) into the polynomial \((1 + x + x^2)^n\):
   \[
   (1 + 1 + 1^2)^n = 3^n
   \]
   This gives the sum of all coefficients:
   \[
   a_0 + a_1 + a_2 + \cdots + a_{2n} = 3^n
   \]

2. Substituting \(x = -1\) into the polynomial \((1 + x + x^2)^n\):
   \[
   (1 + (-1) + (-1)^2)^n = (1 - 1 + 1)^n = 1^n = 1
   \]
   This gives the alternating sum of the coefficients:
   \[
   a_0 - a_1 + a_2 - a_3 + \cdots + a_{2n} = 1
   \]

Next, we add these two results:
\[
(a_0 + a_1 + a_2 + \cdots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \cdots + a_{2n}) = 3^n + 1
\]

Notice that the odd-indexed coefficients cancel out, leaving us with twice the sum of the even-indexed coefficients:
\[
2(a_0 + a_2 + a_4 + \cdots + a_{2n}) = 3^n + 1
\]

Dividing both sides by 2, we obtain the sum of the even-indexed coefficients:
\[
a_0 + a_2 + a_4 + \cdots + a_{2n} = \frac{3^n + 1}{2}
\]

Thus, the sum of the even-indexed coefficients is \(\boxed{\frac{3^n + 1}{2}}\).

System: You are a helpful assistant.


[User] Is it possible for an infinite group \( G \) to have a subgroup \( H \neq G \) which is of finite index in \( G \)? Provide an example.

[Assistant]
Certainly! Here is the clean, final correct solution:

To determine if it is possible for an infinite group \( G \) to have a subgroup \( H \neq G \) which is of finite index in \( G \), we can provide the following example:

Consider the infinite group \( G = \mathbb{Z} \) (the set of all integers under addition). Let \( H = 2\mathbb{Z} \), the set of all even integers.

1. **Verification that \( H \) is a subgroup of \( G \):**
   - \( H \) is non-empty since \( 0 \in 2\mathbb{Z} \).
   - If \( a, b \in 2\mathbb{Z} \), then \( a + b \in 2\mathbb{Z} \) because the sum of two even numbers is even.
   - If \( a \in 2\mathbb{Z} \), then \( -a \in 2\mathbb{Z} \) because the negative of an even number is even.
   Therefore, \( H = 2\mathbb{Z} \) is a subgroup of \( G = \mathbb{Z} \).

2. **Verification that \( H \neq G \):**
   - \( G = \mathbb{Z} \) contains odd integers, but \( H = 2\mathbb{Z} \) does not. For example, \( 1 \in \mathbb{Z} \) but \( 1 \notin 2\mathbb{Z} \).

3. **Verification that \( H \) has finite index in \( G \):**
   - The cosets of \( H \) in \( G \) are \( 0 + H = 2\mathbb{Z} \) and \( 1 + H = 1 + 2\mathbb{Z} \).
   - These are the only distinct cosets because any integer \( n \) can be written as \( n = 2k \) or \( n = 2k + 1 \) for some integer \( k \).
   - Therefore, the index of \( H \) in \( G \) is 2, which is finite.

Thus, \( G = \mathbb{Z} \) is an infinite group, \( H = 2\mathbb{Z} \) is a subgroup of \( G \) with \( H \neq G \), and \( H \) has finite index 2 in \( G \).

\[
\boxed{\text{Yes}}
\]

System: You are a helpful assistant.


[User] How many common points can at least two lines have in a set of 3 plan lines?

[Assistant]
To determine the maximum number of common points where at least two lines intersect in a set of three plane lines, we need to consider the possible configurations of the lines.

1. **Configuration Analysis:**
    - **Case 1: All three lines are parallel.**
        - In this case, no two lines intersect, so there are no common points.
    - **Case 2: Two lines are parallel, and the third line is not.**
        - The third line intersects each of the two parallel lines at two distinct points, resulting in 2 common points.
    - **Case 3: All three lines are concurrent (intersect at a single point).**
        - Here, all three lines intersect at one common point, resulting in 1 common point.
    - **Case 4: The three lines are in general position (no two are parallel, and no three are concurrent).**
        - Each pair of lines intersects at a unique point. Since there are three pairs of lines, there are three distinct intersection points.

2. **Conclusion:**
    - The maximum number of common points where at least two lines intersect is achieved when the three lines are in general position, resulting in three distinct intersection points.

Thus, the maximum number of common points where at least two lines intersect in a set of three plane lines is \(\boxed{3}\).

System: You are a helpful assistant.


[User] Two years ago there were 20 trailer homes on Elm Street with an average age of 18 years. At that time, a group of brand new trailer homes was then added to Elm Street. Today, the average age of all the trailer homes on Elm Street is 14 years. How many new trailer homes were added two years ago?

[Assistant]
To solve the problem, we need to determine how many new trailer homes were added to Elm Street two years ago.

First, let's establish the given information:
- Two years ago, there were 20 trailer homes with an average age of 18 years.
- At that time, a group of brand new trailer homes was added.
- Today, the average age of all trailer homes is 14 years.

Let \( x \) be the number of new trailer homes added two years ago.

### Step-by-Step Solution:

1. **Determine the total age of the original trailer homes two years ago:**
   - Two years ago, the average age of the 20 original trailer homes was 18 years.
   - Therefore, the total age of these 20 trailer homes two years ago was:
     \[
     20 \times 18 = 360 \text{ years}
     \]

2. **Calculate the current total age of the original trailer homes:**
   - Since two years have passed, each of the original 20 trailer homes has aged by 2 years.
   - Thus, the current total age of the original trailer homes is:
     \[
     360 + 20 \times 2 = 360 + 40 = 400 \text{ years}
     \]

3. **Calculate the current total age of the new trailer homes:**
   - The new trailer homes were added two years ago and are brand new at that time, so each is now 2 years old.
   - Therefore, the total age of the \( x \) new trailer homes is:
     \[
     x \times 2 = 2x \text{ years}
     \]

4. **Determine the combined total age of all trailer homes today:**
   - The total number of trailer homes today is \( 20 + x \).
   - The combined total age of all trailer homes today is:
     \[
     400 + 2x \text{ years}
     \]

5. **Set up the equation for the average age:**
   - The average age of all trailer homes today is 14 years.
   - Therefore, we can set up the equation:
     \[
     \frac{400 + 2x}{20 + x} = 14
     \]

6. **Solve the equation for \( x \):**
   - Multiply both sides by \( 20 + x \) to clear the denominator:
     \[
     400 + 2x = 14(20 + x)
     \]
   - Expand the right-hand side:
     \[
     400 + 2x = 280 + 14x
     \]
   - Subtract \( 2x \) from both sides:
     \[
     400 = 280 + 12x
     \]
   - Subtract 280 from both sides:
     \[
     120 = 12x
     \]
   - Divide both sides by 12:
     \[
     x = 10
     \]

Thus, the number of new trailer homes added two years ago is \(\boxed{10}\).

System: You are a helpful assistant.


[User] What is the product of the least common multiple and the greatest common factor of $20$ and $90$?

[Assistant]
To find the product of the least common multiple (LCM) and the greatest common factor (GCF) of 20 and 90, we can use the relationship between LCM and GCF, which states that:

\[
\text{LCM}(a, b) \times \text{GCF}(a, b) = a \times b
\]

First, let's find the GCF of 20 and 90. The factors of 20 are \(1, 2, 4, 5, 10, 20\), and the factors of 90 are \(1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90\). The greatest common factor is the largest number that divides both 20 and 90 without leaving a remainder, which is 10. Therefore, \(\text{GCF}(20, 90) = 10\).

Next, we use the relationship to find the LCM. According to the formula:

\[
\text{LCM}(20, 90) = \frac{20 \times 90}{\text{GCF}(20, 90)} = \frac{20 \times 90}{10} = \frac{1800}{10} = 180
\]

So, \(\text{LCM}(20, 90) = 180\).

Now, we calculate the product of the LCM and the GCF:

\[
\text{LCM}(20, 90) \times \text{GCF}(20, 90) = 180 \times 10 = 1800
\]

Alternatively, using the relationship directly:

\[
20 \times 90 = 1800
\]

Thus, the product of the least common multiple and the greatest common factor of 20 and 90 is:

\[
\boxed{1800}
\]

System: You are a helpful assistant.


[User] In how many ways can a dog trainer buy 8 dogs, all of which are either cocker spaniels, Irish setters, or Russian wolfhounds, if she wants at least one from each category?

[Assistant]
To solve the problem of finding the number of ways a dog trainer can buy 8 dogs, choosing from three breeds (cocker spaniels, Irish setters, and Russian wolfhounds), with the condition that she must have at least one from each category, we can use the stars and bars method with a slight adjustment.

1. **Set Up the Problem:**
   We need to find the number of solutions to the equation \( x + y + z = 8 \) where \( x, y, z \geq 1 \). Here, \( x \) represents the number of cocker spaniels, \( y \) the number of Irish setters, and \( z \) the number of Russian wolfhounds.

2. **Adjust for the Constraint:**
   Since each breed must have at least one dog, we can transform the variables to \( x' = x - 1 \), \( y' = y - 1 \), and \( z' = z - 1 \). This transformation ensures that \( x', y', z' \geq 0 \).

   Substituting these into the original equation:
   \[
   (x' + 1) + (y' + 1) + (z' + 1) = 8
   \]
   Simplifying, we get:
   \[
   x' + y' + z' = 5
   \]

3. **Apply the Stars and Bars Method:**
   The equation \( x' + y' + z' = 5 \) represents the distribution of 5 "stars" (remaining dogs) into 3 "bins" (breeds), where each bin can have zero or more stars. The number of non-negative integer solutions to this equation is given by the binomial coefficient:
   \[
   \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}
   \]

4. **Calculate the Binomial Coefficient:**
   \[
   \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21
   \]

Therefore, the number of ways the dog trainer can buy 8 dogs, ensuring at least one from each breed, is:
\[
\boxed{21}
\]

System: You are a helpful assistant.


[User] In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?

[Assistant]
To determine the total number of people in the line, we need to use the given information that the 1013th person from the left is also the 1010th person from the right.

Let's denote the total number of people in the line by \( N \).

1. The 1013th person from the left implies there are 1012 people to their left.
2. The 1010th person from the right implies there are 1009 people to their right.

Since the person in question is counted in both the left and right counts, the total number of people in the line can be calculated as:
\[ N = \text{(people to the left)} + \text{(the person)} + \text{(people to the right)} \]
\[ N = 1012 + 1 + 1009 \]

Adding these values:
\[ N = 1012 + 1 + 1009 = 2022 \]

Alternatively, we can use the formula:
\[ N = \text{(position from the left)} + \text{(position from the right)} - 1 \]
\[ N = 1013 + 1010 - 1 = 2022 \]

Thus, the total number of people in the line is:
\[
\boxed{2022}
\]

System: You are a helpful assistant.


[User] The 5-digit number $52\,28\square$ is a multiple of 6.  Which digit is represented by $\square$?

[Assistant]
To determine the digit represented by \(\square\) in the 5-digit number \(5228\square\) such that the number is a multiple of 6, we need to ensure the number is divisible by both 2 and 3.

### Step 1: Check for divisibility by 2
A number is divisible by 2 if its last digit is even. Therefore, the digit \(\square\) must be one of \(0, 2, 4, 6, 8\).

### Step 2: Check for divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. The known digits are \(5, 2, 2, 8\). We need to find the sum of these digits and then determine which even digit \(\square\) makes the total sum divisible by 3.

First, calculate the sum of the known digits:
\[
5 + 2 + 2 + 8 = 17
\]

Let \(x\) be the missing digit. The total sum of the digits will be:
\[
17 + x
\]

For \(17 + x\) to be divisible by 3, \(17 + x \equiv 0 \pmod{3}\).

### Step 3: Simplify the congruence
First, find \(17 \mod 3\):
\[
17 \div 3 = 5 \text{ remainder } 2 \quad \Rightarrow \quad 17 \equiv 2 \pmod{3}
\]

Thus, we need:
\[
2 + x \equiv 0 \pmod{3}
\]

Solving for \(x\):
\[
x \equiv -2 \pmod{3} \quad \Rightarrow \quad x \equiv 1 \pmod{3}
\]

### Step 4: Identify the even digit that satisfies \(x \equiv 1 \pmod{3}\)
The even digits are \(0, 2, 4, 6, 8\). We need to find which of these is congruent to 1 modulo 3:
\[
\begin{align*}
0 &\equiv 0 \pmod{3} \\
2 &\equiv 2 \pmod{3} \\
4 &\equiv 1 \pmod{3} \\
6 &\equiv 0 \pmod{3} \\
8 &\equiv 2 \pmod{3}
\end{align*}
\]

The only even digit that satisfies \(x \equiv 1 \pmod{3}\) is \(4\).

### Step 5: Verify the solution
Check the total sum with \(x = 4\):
\[
17 + 4 = 21
\]

Since \(21\) is divisible by 3, the number \(52284\) is divisible by both 2 and 3, and hence by 6.

Thus, the digit represented by \(\square\) is \(\boxed{4}\).

System: You are a helpful assistant.


[User] You and nine friends initially spend $4000$ dollars on tickets to attend a concert. Six friends cancel last minute due to the flu. You and the remaining friends split the original cost of $4000$ dollars equally. What percent of the total cost does each remaining individual have to pay?

[Assistant]
To solve the problem, we need to determine what percentage of the total cost each remaining individual has to pay after six friends cancel.

1. **Initial Setup:**
   - Total number of people initially: \(10\) (you and nine friends).
   - Total cost of tickets: \( \$4000 \).

2. **Number of Remaining People:**
   - Six friends cancel, so the number of remaining people is \(10 - 6 = 4\).

3. **Cost per Remaining Individual:**
   - The total cost of \( \$4000 \) is now split among the 4 remaining people.
   - Each remaining person pays: \(\frac{4000}{4} = 1000\) dollars.

4. **Percentage Calculation:**
   - To find the percentage of the total cost that each remaining individual has to pay, we use the formula for percentage:
     \[
     \text{Percentage} = \left( \frac{\text{Part}}{\text{Whole}} \right) \times 100
     \]
   - Here, the "Part" is the amount each remaining person pays (\( \$1000 \)), and the "Whole" is the total cost (\( \$4000 \)).
   - Therefore, the percentage is:
     \[
     \left( \frac{1000}{4000} \right) \times 100 = 0.25 \times 100 = 25\%
     \]

Thus, each remaining individual has to pay \(\boxed{25\%}\) of the total cost.

System: You are a helpful assistant.


[User] Solve the equation \( e^{1-5x} = 3.5 \).

[Assistant]
To solve the equation \( e^{1-5x} = 3.5 \), we follow these steps:

1. **Take the natural logarithm of both sides:**
   \[
   \ln(e^{1-5x}) = \ln(3.5)
   \]

2. **Simplify the left side using the property of logarithms \(\ln(e^y) = y\):**
   \[
   1 - 5x = \ln(3.5)
   \]

3. **Isolate the term involving \(x\):**
   \[
   -5x = \ln(3.5) - 1
   \]

4. **Solve for \(x\) by dividing both sides by \(-5\):**
   \[
   x = \frac{\ln(3.5) - 1}{-5}
   \]

5. **Simplify the expression:**
   \[
   x = \frac{1 - \ln(3.5)}{5}
   \]

Thus, the exact solution to the equation \( e^{1-5x} = 3.5 \) is:
\[
\boxed{\frac{1 - \ln(3.5)}{5}}
\]

System: You are a helpful assistant.


[User] Factor the following expression: $37a^2 +111a$.

[Assistant]
To factor the expression \(37a^2 + 111a\), we start by identifying the greatest common factor (GCF) of the terms.

1. **Identify the GCF of the coefficients:**
   - The coefficients are 37 and 111.
   - Since 37 is a prime number, its factors are 1 and 37.
   - We check if 37 divides 111: \(111 \div 37 = 3\), so 37 is a factor of 111.
   - Therefore, the GCF of 37 and 111 is 37.

2. **Identify the GCF of the variable part:**
   - The terms are \(37a^2\) and \(111a\).
   - The lowest power of \(a\) in both terms is \(a\).
   - Therefore, the GCF of the variable part is \(a\).

3. **Combine the GCF of the coefficients and the variables:**
   - The GCF of \(37a^2\) and \(111a\) is \(37a\).

4. **Factor out the GCF from each term:**
   - Divide \(37a^2\) by \(37a\): \(\frac{37a^2}{37a} = a\).
   - Divide \(111a\) by \(37a\): \(\frac{111a}{37a} = 3\).

5. **Write the expression in factored form:**
   - Factoring out \(37a\) from \(37a^2 + 111a\) gives \(37a(a + 3)\).

6. **Verify the factored form:**
   - Distribute \(37a\) back into \((a + 3)\):
     \[
     37a \cdot a = 37a^2
     \]
     \[
     37a \cdot 3 = 111a
     \]
   - Adding these results gives \(37a^2 + 111a\), which matches the original expression.

Therefore, the fully factored form of the expression is:

\[
\boxed{37a(a + 3)}
\]

System: You are a helpful assistant.


[User] Evaluate the integral $\int_{0}^{t} sgn(x) \ dx$, where $sgn(x) = -1$ for $x < 0$, $sgn(0) = 0$, and $sgn(x) = 1$ for $x > 0$.

[Assistant]
To evaluate the integral \(\int_{0}^{t} \text{sgn}(x) \, dx\), we need to consider the behavior of the signum function \(\text{sgn}(x)\), which is defined as:
\[
\text{sgn}(x) =
\begin{cases}
-1 & \text{if } x < 0, \\
0 & \text{if } x = 0, \\
1 & \text{if } x > 0.
\end{cases}
\]

We will evaluate the integral for three cases: \(t > 0\), \(t = 0\), and \(t < 0\).

1. **Case 1: \(t > 0\)**

   When \(t > 0\), the function \(\text{sgn}(x)\) is equal to 1 for all \(x\) in the interval \([0, t]\). Thus, the integral becomes:
   \[
   \int_{0}^{t} \text{sgn}(x) \, dx = \int_{0}^{t} 1 \, dx = [x]_{0}^{t} = t - 0 = t.
   \]

2. **Case 2: \(t = 0\)**

   When \(t = 0\), the integral is over an interval of length 0, so:
   \[
   \int_{0}^{0} \text{sgn}(x) \, dx = 0.
   \]

3. **Case 3: \(t < 0\)**

   When \(t < 0\), the integral from 0 to \(t\) can be rewritten as the negative of the integral from \(t\) to 0:
   \[
   \int_{0}^{t} \text{sgn}(x) \, dx = - \int_{t}^{0} \text{sgn}(x) \, dx.
   \]
   In the interval \([t, 0]\), \(\text{sgn}(x) = -1\) for \(x\) in \((t, 0)\). Therefore:
   \[
   \int_{t}^{0} \text{sgn}(x) \, dx = \int_{t}^{0} (-1) \, dx = [-x]_{t}^{0} = 0 - (-t) = t.
   \]
   Consequently:
   \[
   \int_{0}^{t} \text{sgn}(x) \, dx = -t.
   \]
   Since \(t\) is negative, \(-t\) is the absolute value of \(t\):
   \[
   -t = |t|.
   \]

Combining all cases, we find that the integral \(\int_{0}^{t} \text{sgn}(x) \, dx\) is equal to \(|t|\) regardless of the sign of \(t\). Therefore, the final result is:
\[
\boxed{|t|}
\]

System: You are a helpful assistant.


[User] At a birthday party, everyone who gets an invitation card can come alone or bring a friend. Everyone invited was present at the event. The ratio of the number of invitation cards to the number of guests attending is $3:5$. What is the ratio of the number of invitation cards to the number of guests who bring friends?

[Assistant]
To solve the problem, we need to determine the ratio of the number of invitation cards to the number of guests who bring friends, given that the ratio of invitation cards to the total number of guests attending is \(3:5\).

Let's denote the number of invitation cards by \(3x\) and the total number of guests by \(5x\). Each invitation card can result in either 1 guest (if the invitee comes alone) or 2 guests (if the invitee brings a friend).

Let \(y\) be the number of invitees who brought a friend. Then, the number of invitees who came alone is \(3x - y\). The total number of guests can be expressed as:
\[
(3x - y) \cdot 1 + y \cdot 2 = 5x
\]
Simplifying the equation:
\[
3x - y + 2y = 5x \implies 3x + y = 5x \implies y = 2x
\]
Thus, the number of guests who brought friends is \(y = 2x\). These guests are the invitees who brought a friend, and each such invitee is a guest themselves.

The ratio of the number of invitation cards to the number of guests who brought friends is:
\[
3x : 2x = 3 : 2
\]

Therefore, the ratio of the number of invitation cards to the number of guests who bring friends is \(\boxed{3:2}\).

System: You are a helpful assistant.


[User] Find the number of combinations of a given length \( k \) that use elements from a set of \( n \) elements, allowing repetition.

[Assistant]
To find the number of combinations of length \( k \) from a set of \( n \) elements, allowing repetition, we use the concept of combinations with repetition. The formula for the number of combinations with repetition is given by the binomial coefficient \(\binom{n + k - 1}{k}\).

Here is the detailed reasoning:

1. **Understanding the Problem**: We need to determine the number of ways to choose \( k \) elements from a set of \( n \) elements, where repetition of elements is allowed, and the order of selection does not matter.

2. **Stars and Bars Theorem**: This problem can be translated into a "stars and bars" problem. Imagine we have \( k \) indistinguishable "stars" (representing the elements we are choosing) and \( n-1 \) "bars" that separate these stars into \( n \) distinct groups (each group corresponding to one of the \( n \) elements).

3. **Counting the Arrangements**: The total number of positions to place the stars and bars is \( n + k - 1 \). We need to choose \( k \) positions out of these \( n + k - 1 \) positions to place the stars (or equivalently, choose \( n-1 \) positions to place the bars). This is a combinatorial problem of choosing \( k \) positions from \( n + k - 1 \) positions, which is given by the binomial coefficient \(\binom{n + k - 1}{k}\).

4. **Formula for Combinations with Repetition**: The number of combinations of \( k \) elements from \( n \) elements with repetition allowed is therefore \(\binom{n + k - 1}{k}\).

Thus, the number of combinations of length \( k \) from a set of \( n \) elements with repetition allowed is \(\boxed{\binom{n + k - 1}{k}}\).

System: You are a helpful assistant.


[User] Evaluate the expression \[ \frac{a+2}{a+1} \cdot \frac{b-1}{b-2} \cdot \frac{c + 8}{c+6} , \] given that $c = b-10$, $b = a+2$, $a = 4$, and none of the denominators are zero.

[Assistant]
To evaluate the expression \(\frac{a+2}{a+1} \cdot \frac{b-1}{b-2} \cdot \frac{c + 8}{c+6}\) given \(c = b-10\), \(b = a+2\), and \(a = 4\), we start by substituting the given values into the expression.

First, we find the values of \(b\) and \(c\):
\[
a = 4
\]
\[
b = a + 2 = 4 + 2 = 6
\]
\[
c = b - 10 = 6 - 10 = -4
\]

Next, we substitute these values into the expression:
\[
\frac{a+2}{a+1} \cdot \frac{b-1}{b-2} \cdot \frac{c + 8}{c+6}
\]
\[
= \frac{4+2}{4+1} \cdot \frac{6-1}{6-2} \cdot \frac{-4 + 8}{-4 + 6}
\]
\[
= \frac{6}{5} \cdot \frac{5}{4} \cdot \frac{4}{2}
\]

Now, we simplify each fraction:
\[
\frac{6}{5} \cdot \frac{5}{4} \cdot \frac{4}{2}
\]

We can simplify the expression by canceling common factors:
\[
= \frac{6 \cdot 5 \cdot 4}{5 \cdot 4 \cdot 2}
\]
\[
= \frac{6 \cdot \cancel{5} \cdot \cancel{4}}{\cancel{5} \cdot \cancel{4} \cdot 2}
\]
\[
= \frac{6}{2}
\]
\[
= 3
\]

Thus, the value of the expression is \(\boxed{3}\).

System: You are a helpful assistant.


[User] Molly flips a fair coin five times, and she is very surprised to flip a head each time. What is the probability she will flip a tail on her next flip of the coin? Express your answer as a common fraction.

[Assistant]
To solve the problem of finding the probability that Molly will flip a tail on her next flip of a fair coin, we need to consider the following key points:

1. **Independence of Coin Flips**: Each flip of a fair coin is an independent event. This means that the outcome of one flip does not affect the outcome of any other flip.
2. **Fair Coin**: A fair coin has two sides, heads and tails, and each side has an equal probability of landing face up. Therefore, the probability of getting heads or tails on any single flip is \( \frac{1}{2} \).

Given that Molly has already flipped the coin five times and obtained heads each time, we need to determine the probability of getting a tail on the next flip. Since each flip is independent, the previous outcomes (five heads in a row) do not influence the probability of the next flip.

Thus, the probability of flipping a tail on the next flip is simply the probability of getting tails on a single flip of a fair coin, which is \( \frac{1}{2} \).

Therefore, the probability that Molly will flip a tail on her next flip is \(\boxed{\frac{1}{2}}\).

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 the reciprocal of \( x+1 \) is \( x-1 \), determine \( x \).

[Assistant]
To solve the problem where the reciprocal of \( x + 1 \) is \( x - 1 \), we start by setting up the equation:

\[
\frac{1}{x + 1} = x - 1
\]

Next, we eliminate the fraction by multiplying both sides of the equation by \( x + 1 \):

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

Using the difference of squares formula, we simplify the right-hand side:

\[
1 = x^2 - 1
\]

To isolate \( x^2 \), we add 1 to both sides of the equation:

\[
x^2 = 2
\]

Taking the square root of both sides, we get:

\[
x = \pm \sqrt{2}
\]

Now, we need to verify that both solutions \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) satisfy the original equation without causing any division by zero.

1. **Verification for \( x = \sqrt{2} \):**

   Substitute \( x = \sqrt{2} \) into the original equation:

   \[
   \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1
   \]

   To verify, we rationalize the left-hand side:

   \[
   \frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1
   \]

   This confirms that \( x = \sqrt{2} \) is a valid solution.

2. **Verification for \( x = -\sqrt{2} \):**

   Substitute \( x = -\sqrt{2} \) into the original equation:

   \[
   \frac{1}{-\sqrt{2} + 1} = -\sqrt{2} - 1
   \]

   To verify, we rationalize the left-hand side:

   \[
   \frac{1}{-\sqrt{2} + 1} \cdot \frac{-\sqrt{2} - 1}{-\sqrt{2} - 1} = \frac{-\sqrt{2} - 1}{(-\sqrt{2})^2 - 1^2} = \frac{-\sqrt{2} - 1}{2 - 1} = -\sqrt{2} - 1
   \]

   This confirms that \( x = -\sqrt{2} \) is also a valid solution.

Thus, the solutions to the equation are:

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

System: You are a helpful assistant.


[User] The sum of two numbers is $45$. Their difference is $3$. What is the lesser of the two numbers?

[Assistant]
To solve the problem where the sum of two numbers is 45 and their difference is 3, we can use a system of equations. Let the two numbers be \( x \) (the larger number) and \( y \) (the smaller number). We can set up the following system of equations based on the given information:

\[
\begin{cases}
x + y = 45 \\
x - y = 3
\end{cases}
\]

First, we add the two equations together to eliminate \( y \):

\[
(x + y) + (x - y) = 45 + 3
\]

This simplifies to:

\[
2x = 48
\]

Next, we solve for \( x \):

\[
x = \frac{48}{2} = 24
\]

Now that we have \( x = 24 \), we substitute this value back into the first equation to find \( y \):

\[
24 + y = 45
\]

Solving for \( y \):

\[
y = 45 - 24 = 21
\]

Thus, the two numbers are 24 and 21. The problem asks for the lesser of the two numbers, which is \( y \).

Therefore, the lesser of the two numbers is:

\[
\boxed{21}
\]

System: You are a helpful assistant.


[User] Expand the product ${3(x+4)(x+5)}$.

[Assistant]
To expand the product \(3(x+4)(x+5)\), we will follow these steps:

1. First, expand the binomials \((x+4)\) and \((x+5)\) using the distributive property (FOIL method).
2. Combine like terms.
3. Multiply the resulting polynomial by 3.

Let's begin with the binomials \((x+4)\) and \((x+5)\):

\[
(x+4)(x+5)
\]

Using the FOIL method:

- **First terms**: \(x \cdot x = x^2\)
- **Outer terms**: \(x \cdot 5 = 5x\)
- **Inner terms**: \(4 \cdot x = 4x\)
- **Last terms**: \(4 \cdot 5 = 20\)

Adding these together:

\[
x^2 + 5x + 4x + 20
\]

Combine the like terms \(5x\) and \(4x\):

\[
x^2 + 9x + 20
\]

Now, we need to multiply this result by 3:

\[
3(x^2 + 9x + 20)
\]

Distribute the 3 to each term inside the parentheses:

\[
3 \cdot x^2 = 3x^2
\]
\[
3 \cdot 9x = 27x
\]
\[
3 \cdot 20 = 60
\]

Putting it all together, the expanded form is:

\[
3x^2 + 27x + 60
\]

Thus, the final answer is:

\[
\boxed{3x^2 + 27x + 60}
\]

System: You are a helpful assistant.


[User] Integrate \(\frac{\sin x}{x}\) with respect to \(x\).

[Assistant]
The integral of \(\frac{\sin x}{x}\) with respect to \(x\) cannot be expressed in terms of elementary functions. Instead, it is represented by a special function known as the sine integral, denoted as \(\text{Si}(x)\).

The sine integral function \(\text{Si}(x)\) is defined as:

\[
\text{Si}(x) = \int_0^x \frac{\sin t}{t} \, dt
\]

Given the problem, we need to find the indefinite integral of \(\frac{\sin x}{x}\):

\[
\int \frac{\sin x}{x} \, dx
\]

Since \(\frac{\sin x}{x}\) does not have an elementary antiderivative, we express the result in terms of the sine integral function \(\text{Si}(x)\):

\[
\int \frac{\sin x}{x} \, dx = \text{Si}(x) + C
\]

where \(C\) is the constant of integration.

Therefore, the solution is:

\[
\boxed{\text{Si}(x) + C}
\]

System: You are a helpful assistant.


[User] Triangle $ABC$ is isosceles with angle $A$ congruent to angle $B$. The measure of angle $C$ is 30 degrees more than the measure of angle $A$. What is the number of degrees in the measure of angle $C$?

[Assistant]
To find the measure of angle \( C \) in the isosceles triangle \( ABC \) where \(\angle A \cong \angle B\) and \(\angle C\) is 30 degrees more than \(\angle A\), we proceed as follows:

1. Let the measure of \(\angle A\) be \( x \) degrees. Since \(\angle A \cong \angle B\), the measure of \(\angle B\) is also \( x \) degrees.
2. The measure of \(\angle C\) is given as \( x + 30 \) degrees.
3. The sum of the interior angles in a triangle is 180 degrees. Therefore, we can write the equation:
   \[
   \angle A + \angle B + \angle C = 180
   \]
   Substituting the given values, we get:
   \[
   x + x + (x + 30) = 180
   \]
4. Simplifying the equation:
   \[
   3x + 30 = 180
   \]
5. Subtract 30 from both sides of the equation:
   \[
   3x = 150
   \]
6. Divide both sides by 3:
   \[
   x = 50
   \]
7. Therefore, the measure of \(\angle C\) is:
   \[
   x + 30 = 50 + 30 = 80
   \]

Thus, the measure of angle \( C \) is \(\boxed{80}\) degrees.

System: You are a helpful assistant.


[User] John recently bought a used car for $\$5000$ for his pizza delivery job. He gets $\$10$ for each pizza he delivers, but he has to spend $\$3$ on gas for each pizza he delivers. What is the minimum whole number of pizzas John must deliver in order to earn back the money he spent on the car he bought?

[Assistant]
To determine the minimum whole number of pizzas John must deliver to earn back the \$5000 he spent on the car, we need to calculate his net profit per pizza and then find out how many pizzas he needs to deliver to reach at least \$5000 in profit.

1. **Calculate the net profit per pizza:**
   - John earns \$10 for each pizza he delivers.
   - He spends \$3 on gas for each pizza.
   - Therefore, his net profit per pizza is:
     \[
     10 - 3 = 7 \text{ dollars per pizza}
     \]

2. **Determine the number of pizzas needed:**
   - John needs to earn back \$5000.
   - Let \( x \) be the number of pizzas he needs to deliver.
   - The total profit from delivering \( x \) pizzas is \( 7x \).
   - We need \( 7x \geq 5000 \).

3. **Solve for \( x \):**
   \[
   7x \geq 5000
   \]
   \[
   x \geq \frac{5000}{7}
   \]
   \[
   x \geq 714.2857
   \]

4. **Round up to the nearest whole number:**
   - Since John cannot deliver a fraction of a pizza, we round up 714.2857 to the next whole number.
   - Therefore, \( x = 715 \).

Thus, John must deliver at least \(\boxed{715}\) pizzas to earn back the \$5000 he spent on the car.

System: You are a helpful assistant.


[User] Khali has to shovel snow off the sidewalk in front of his house. The sidewalk is 20 feet long and 2 feet wide. If the snow is $\frac{1}{2}$ foot deep, how many cubic feet of snow does Khali have to shovel off the sidewalk?

[Assistant]
To determine the volume of snow Khali has to shovel, we need to calculate the volume of a rectangular prism, which is given by the formula:

\[
\text{Volume} = \text{length} \times \text{width} \times \text{depth}
\]

Given the dimensions of the sidewalk and the depth of the snow:
- Length = 20 feet
- Width = 2 feet
- Depth = \(\frac{1}{2}\) foot

Substituting these values into the formula, we get:

\[
\text{Volume} = 20 \, \text{ft} \times 2 \, \text{ft} \times \frac{1}{2} \, \text{ft}
\]

First, we multiply the length and the width:

\[
20 \, \text{ft} \times 2 \, \text{ft} = 40 \, \text{ft}^2
\]

Next, we multiply the result by the depth:

\[
40 \, \text{ft}^2 \times \frac{1}{2} \, \text{ft} = 20 \, \text{cubic feet}
\]

Thus, the volume of snow Khali has to shovel is:

\[
\boxed{20}
\] cubic feet.

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] Solve for natural numbers \( n \) in the equation \( n^3 - n = n! \).

[Assistant]
To solve the equation \( n^3 - n = n! \) for natural numbers \( n \), we will proceed step-by-step.

First, note that natural numbers are positive integers, i.e., \( n = 1, 2, 3, \ldots \).

1. **Check small values of \( n \):**

   - For \( n = 1 \):
     \[
     n^3 - n = 1^3 - 1 = 0
     \]
     \[
     n! = 1! = 1
     \]
     Since \( 0 \neq 1 \), \( n = 1 \) is not a solution.

   - For \( n = 2 \):
     \[
     n^3 - n = 2^3 - 2 = 8 - 2 = 6
     \]
     \[
     n! = 2! = 2
     \]
     Since \( 6 \neq 2 \), \( n = 2 \) is not a solution.

   - For \( n = 3 \):
     \[
     n^3 - n = 3^3 - 3 = 27 - 3 = 24
     \]
     \[
     n! = 3! = 6
     \]
     Since \( 24 \neq 6 \), \( n = 3 \) is not a solution.

   - For \( n = 4 \):
     \[
     n^3 - n = 4^3 - 4 = 64 - 4 = 60
     \]
     \[
     n! = 4! = 24
     \]
     Since \( 60 \neq 24 \), \( n = 4 \) is not a solution.

   - For \( n = 5 \):
     \[
     n^3 - n = 5^3 - 5 = 125 - 5 = 120
     \]
     \[
     n! = 5! = 120
     \]
     Since \( 120 = 120 \), \( n = 5 \) is a solution.

2. **Analyze the behavior for \( n \geq 6 \):**

   - For \( n = 6 \):
     \[
     n^3 - n = 6^3 - 6 = 216 - 6 = 210
     \]
     \[
     n! = 6! = 720
     \]
     Since \( 210 < 720 \), \( n = 6 \) is not a solution.

   - For \( n = 7 \):
     \[
     n^3 - n = 7^3 - 7 = 343 - 7 = 336
     \]
     \[
     n! = 7! = 5040
     \]
     Since \( 336 < 5040 \), \( n = 7 \) is not a solution.

   - Generally, for \( n \geq 6 \), \( n! \) grows much faster than \( n^3 - n \). Therefore, \( n^3 - n \) will always be less than \( n! \) for \( n \geq 6 \).

Thus, the only natural number \( n \) that satisfies the equation \( n^3 - n = n! \) is \( n = 5 \).

\[
\boxed{5}
\]

System: You are a helpful assistant.


[User] Let \( F(x) = x^{\cos x + \sin x} \). Find \( F'(1) \).

[Assistant]
To find \( F'(1) \) where \( F(x) = x^{\cos x + \sin x} \), we use logarithmic differentiation.

1. Let \( y = x^{\cos x + \sin x} \). Taking the natural logarithm of both sides, we get:
   \[
   \ln y = (\cos x + \sin x) \ln x
   \]

2. Differentiate both sides with respect to \( x \):
   \[
   \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ (\cos x + \sin x) \ln x \right]
   \]

3. Use the product rule on the right-hand side. Let \( u = \cos x + \sin x \) and \( v = \ln x \):
   \[
   \frac{d}{dx} [uv] = u'v + uv'
   \]

4. Calculate \( u' \) and \( v' \):
   \[
   u' = -\sin x + \cos x, \quad v' = \frac{1}{x}
   \]

5. Applying the product rule:
   \[
   \frac{d}{dx} [(\cos x + \sin x) \ln x] = (-\sin x + \cos x) \ln x + (\cos x + \sin x) \frac{1}{x}
   \]

6. Substitute back into the differentiated equation:
   \[
   \frac{1}{y} \frac{dy}{dx} = (-\sin x + \cos x) \ln x + \frac{\cos x + \sin x}{x}
   \]

7. Multiply both sides by \( y \):
   \[
   \frac{dy}{dx} = y \left[ (\cos x - \sin x) \ln x + \frac{\cos x + \sin x}{x} \right]
   \]

8. Since \( y = x^{\cos x + \sin x} \), we have:
   \[
   \frac{dy}{dx} = x^{\cos x + \sin x} \left[ (\cos x - \sin x) \ln x + \frac{\cos x + \sin x}{x} \right]
   \]

9. Evaluate at \( x = 1 \):
   - \( x^{\cos x + \sin x} \) at \( x = 1 \) is \( 1^{\cos 1 + \sin 1} = 1 \)
   - \( \ln 1 = 0 \), so the term \( (\cos 1 - \sin 1) \ln 1 \) becomes 0
   - The remaining term is \( \frac{\cos 1 + \sin 1}{1} \)

10. Therefore, the derivative at \( x = 1 \) is:
    \[
    F'(1) = \cos 1 + \sin 1
    \]

The final answer is:
\[
\boxed{\cos 1 + \sin 1}
\]

System: You are a helpful assistant.


[User] Find $\ln{-1}$.

[Assistant]
To find \(\ln(-1)\), we need to use the properties of complex logarithms. The natural logarithm function \(\ln(x)\) is not defined for negative real numbers in the real number system, but it can be extended to the complex plane.

Euler's formula states that \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). For \(\theta = \pi\), we have:
\[ e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1. \]

Taking the natural logarithm of both sides, we get:
\[ \ln(e^{i\pi}) = \ln(-1). \]

Since \(\ln(e^z) = z\) for any complex number \(z\), we have:
\[ \ln(e^{i\pi}) = i\pi. \]

Therefore,
\[ \ln(-1) = i\pi. \]

However, the complex logarithm is multi-valued due to the periodicity of the angle in the complex plane. Specifically, \(e^{i\theta} = e^{i(\theta + 2k\pi)}\) for any integer \(k\). Thus, the general form of \(\ln(-1)\) is:
\[ \ln(-1) = i\pi + 2k\pi i \]
for any integer \(k\).

The principal value of the complex logarithm, which is the value typically used unless otherwise specified, is obtained by choosing \(k = 0\). Therefore, the principal value of \(\ln(-1)\) is:
\[ \ln(-1) = i\pi. \]

Thus, the final answer is:
\[ \boxed{i\pi}. \]

System: You are a helpful assistant.


[User] A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?

[Assistant]
To solve the problem, we need to determine the marked price \( M \) as a percentage of the list price \( L \) such that the merchant can give a 20% discount on the marked price and still make a 25% profit on the selling price.

1. **Determine the cost price \( C \):**
   The merchant buys goods at 25% off the list price \( L \). Therefore, the cost price \( C \) is:
   \[
   C = 0.75L
   \]

2. **Express the selling price \( S \) in terms of the marked price \( M \):**
   The merchant gives a 20% discount on the marked price \( M \), so the selling price \( S \) is:
   \[
   S = 0.8M
   \]

3. **Relate the cost price \( C \) to the selling price \( S \):**
   The merchant wants to make a 25% profit on the selling price \( S \). This means the cost price \( C \) is 75% of the selling price \( S \):
   \[
   C = 0.75S
   \]

4. **Substitute the cost price \( C \) into the equation:**
   From step 1, we know \( C = 0.75L \). Substituting this into the equation \( C = 0.75S \) gives:
   \[
   0.75L = 0.75S
   \]
   Dividing both sides by 0.75, we get:
   \[
   L = S
   \]

5. **Relate the selling price \( S \) to the marked price \( M \):**
   From step 2, we know \( S = 0.8M \). Since \( L = S \), we have:
   \[
   L = 0.8M
   \]
   Solving for \( M \), we get:
   \[
   M = \frac{L}{0.8} = 1.25L
   \]

Thus, the marked price \( M \) must be 125% of the list price \( L \).

The final answer is:
\[
\boxed{125\%}
\]

System: You are a helpful assistant.


[User] Evaluate the integral \(\int e^{1/z} \, dz\) along the boundary of the rectangle centered at \(1 + 2i\) with width 5 and height 6. The boundary is divided into four contours.

[Assistant]
To evaluate the integral \(\int e^{1/z} \, dz\) along the boundary of the rectangle centered at \(1 + 2i\) with width 5 and height 6, we proceed as follows:

1. **Determine the bounds of the rectangle:**
   - The center of the rectangle is at \(1 + 2i\).
   - The width is 5, so the horizontal sides range from \(1 - 2.5 = -1.5\) to \(1 + 2.5 = 3.5\).
   - The height is 6, so the vertical sides range from \(2i - 3i = -i\) to \(2i + 3i = 5i\).
   - Therefore, the rectangle extends from \(-1.5\) to \(3.5\) on the real axis and from \(-i\) to \(5i\) on the imaginary axis.

2. **Check if the singularity is inside the rectangle:**
   - The function \(e^{1/z}\) has an essential singularity at \(z = 0\).
   - The point \(z = 0\) (which is \(0 + 0i\)) lies within the rectangle since \(-1.5 \leq 0 \leq 3.5\) and \(-1 \leq 0 \leq 5\).

3. **Apply the Residue Theorem:**
   - The Residue Theorem states that for a function \(f(z)\) analytic inside and on a simple closed contour \(C\), except for isolated singularities inside \(C\), the integral \(\oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k)\), where the sum is over all residues of \(f(z)\) inside \(C\).
   - Here, the only singularity inside the rectangle is at \(z = 0\).

4. **Compute the residue at \(z = 0\):**
   - The Laurent series expansion of \(e^{1/z}\) around \(z = 0\) is \(\sum_{n=0}^{\infty} \frac{1}{n! z^n}\).
   - The residue is the coefficient of the \(\frac{1}{z}\) term, which corresponds to \(n = 1\).
   - The coefficient is \(\frac{1}{1!} = 1\).

5. **Calculate the integral:**
   - By the Residue Theorem, \(\oint_C e^{1/z} \, dz = 2\pi i \times \text{Res}(e^{1/z}, 0) = 2\pi i \times 1 = 2\pi i\).

Therefore, the value of the integral is \(\boxed{2\pi i}\).

System: You are a helpful assistant.


[User] Evaluate the sum \( \sum_{i=0}^{\infty} \frac{(-1)^i}{2i+1} \).

[Assistant]
To evaluate the sum \( \sum_{i=0}^{\infty} \frac{(-1)^i}{2i+1} \), we start by recognizing that this series is the Taylor series expansion of the arctangent function at \( x = 1 \).

The Taylor series expansion for \( \arctan(x) \) around \( x = 0 \) is given by:
\[
\arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}
\]
for \( |x| \leq 1 \).

By setting \( x = 1 \) in the series, we get:
\[
\arctan(1) = \sum_{n=0}^{\infty} \frac{(-1)^n \cdot 1^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}
\]

We know that \( \arctan(1) = \frac{\pi}{4} \) because \( \tan\left(\frac{\pi}{4}\right) = 1 \).

Thus, the series \( \sum_{i=0}^{\infty} \frac{(-1)^i}{2i+1} \) converges to \( \frac{\pi}{4} \).

To confirm convergence, we use the alternating series test. The terms \( \frac{1}{2n+1} \) decrease monotonically to zero as \( n \to \infty \). Therefore, the series converges.

Therefore, the sum is:
\[
\boxed{\frac{\pi}{4}}
\]

System: You are a helpful assistant.


[User] How many perfect squares are two-digit and divisible by $3?$

[Assistant]
To determine how many two-digit perfect squares are divisible by 3, we first identify the range of integers whose squares are two-digit numbers. The smallest integer whose square is a two-digit number is 4 (since \(4^2 = 16\)), and the largest is 9 (since \(9^2 = 81\)). Therefore, the two-digit perfect squares are:

\[16, 25, 36, 49, 64, 81\]

Next, we need to check which of these numbers are divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. We check each number:

- \(16: 1 + 6 = 7\) (not divisible by 3)
- \(25: 2 + 5 = 7\) (not divisible by 3)
- \(36: 3 + 6 = 9\) (divisible by 3)
- \(49: 4 + 9 = 13\) (not divisible by 3)
- \(64: 6 + 4 = 10\) (not divisible by 3)
- \(81: 8 + 1 = 9\) (divisible by 3)

Thus, the two-digit perfect squares that are divisible by 3 are 36 and 81.

Alternatively, since a perfect square divisible by 3 must be the square of a multiple of 3, we consider the multiples of 3 within the range of square roots (4 to 9). The multiples of 3 in this range are 6 and 9. Their squares are:

\[6^2 = 36\]
\[9^2 = 81\]

Thus, the two-digit perfect squares divisible by 3 are 36 and 81.

Therefore, the number of two-digit perfect squares divisible by 3 is \(\boxed{2}\).

System: You are a helpful assistant.