[User] How can you add eight 8's to get the number 1000?

[Assistant]
To solve the problem of adding eight 8's to get the number 1000, we need to creatively combine the digits into multi-digit numbers. Here is the clean and detailed solution:

1. Start by considering the largest possible multi-digit number using the digit 8. The largest such number we can form is 888, which uses three 8's.
2. Next, consider the remaining five 8's. We can form another multi-digit number using two of these 8's, which is 88.
3. Now, we have used five 8's (three in 888 and two in 88), leaving us with three 8's.
4. Add the numbers together:
   \[
   888 + 88 = 976
   \]
5. Add the remaining three 8's to 976:
   \[
   976 + 8 + 8 + 8 = 976 + 24 = 1000
   \]
6. Verify the total number of 8's used:
   - 888 uses 3 eights.
   - 88 uses 2 eights.
   - 8 + 8 + 8 uses 3 eights.
   - Total: 3 + 2 + 3 = 8 eights.

Thus, the final solution is:
\[
888 + 88 + 8 + 8 + 8 = 1000
\]

So the answer is:
\[
\boxed{888 + 88 + 8 + 8 + 8}
\]

System: You are a helpful assistant.


[User] Identify the pattern and find the next term in the series: 11, 13, 18, 28, 45, 71, ...

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 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] Find the number of possible solutions to the Diophantine equation \( w + x + y + z = n \).

[Assistant]
To find the number of possible solutions to the Diophantine equation \( w + x + y + z = n \) where \( w, x, y, z \) are non-negative integers, we can use the stars and bars theorem. This theorem provides a way to determine the number of ways to distribute \( n \) identical items (stars) into \( k \) distinct groups (bins).

In this problem, we have 4 variables, so \( k = 4 \). The formula for the number of non-negative integer solutions to the equation \( x_1 + x_2 + \cdots + x_k = n \) is given by the binomial coefficient:

\[
\binom{n + k - 1}{k - 1}
\]

For our specific equation \( w + x + y + z = n \), we substitute \( k = 4 \):

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

Thus, the number of non-negative integer solutions to the equation \( w + x + y + z = n \) is:

\[
\boxed{\binom{n + 3}{3}}
\]

System: You are a helpful assistant.


[User] In a business school, 75% of students adopted marketing, 62% adopted finance, and 88% adopted HR as electives. What is the minimum number of students specializing in all three subjects?

[Assistant]
To determine the minimum number of students specializing in all three subjects (marketing, finance, and HR), we use the principle of inclusion-exclusion for three sets. The formula for the minimum overlap is:

\[
\text{Minimum number of students in all three} = M + F + H - 2 \times 100\%
\]

where \(M\), \(F\), and \(H\) are the percentages of students in marketing, finance, and HR respectively. Substituting the given percentages:

\[
M = 75\%, \quad F = 62\%, \quad H = 88\%
\]

We calculate the minimum overlap as follows:

\[
75\% + 62\% + 88\% - 2 \times 100\% = 225\% - 200\% = 25\%
\]

Assuming the total number of students is 100 (a common base for percentage problems), the minimum number of students specializing in all three subjects is 25% of 100, which is 25.

Thus, the minimum number of students specializing in all three subjects is:

\[
\boxed{25}
\]

System: You are a helpful assistant.


[User] The cost of 1 piece of gum is 1 cent.  What is the cost of 1000 pieces of gum, in dollars?

[Assistant]
To determine the cost of 1000 pieces of gum in dollars, we start by noting that each piece of gum costs 1 cent. Therefore, the cost of 1000 pieces of gum is calculated as follows:

1. Calculate the total cost in cents:
   \[
   1000 \text{ pieces} \times 1 \text{ cent per piece} = 1000 \text{ cents}
   \]

2. Convert the total cost from cents to dollars. Since 100 cents is equivalent to 1 dollar:
   \[
   1000 \text{ cents} \div 100 = 10 \text{ dollars}
   \]

Thus, the cost of 1000 pieces of gum is \(\boxed{10}\) dollars.

System: You are a helpful assistant.


[User] In a clothing store, a sweater that was originally priced at $73 was marked down and sold for $54.75. What was the percent amount of the markdown?

[Assistant]
To determine the percent markdown on the sweater, we start by calculating the amount of the markdown in dollars. The original price of the sweater is $73, and it was sold for $54.75. The markdown amount is given by the difference between the original price and the sale price:

\[
73 - 54.75 = 18.25
\]

Next, we need to find the percentage of the markdown. This is done by dividing the markdown amount by the original price and then multiplying by 100:

\[
\left( \frac{18.25}{73} \right) \times 100
\]

First, we perform the division:

\[
\frac{18.25}{73} = 0.25
\]

Then, we multiply by 100 to convert the decimal to a percentage:

\[
0.25 \times 100 = 25
\]

Thus, the percent amount of the markdown is:

\[
\boxed{25\%}
\]

System: You are a helpful assistant.


[User] In how many ways can the letters of the word "PERMUTATIONS" be arranged if the words start with P and end with S?

[Assistant]
To solve the problem of finding the number of ways to arrange the letters of the word "PERMUTATIONS" such that the arrangement starts with P and ends with S, we can follow these steps:

1. **Identify the total number of letters and duplicates:**
   The word "PERMUTATIONS" has 12 letters, with the letter 'T' appearing twice.

2. **Fix the first and last positions:**
   Since the word must start with P and end with S, we fix these two letters at the first and last positions, respectively. This leaves us with 10 letters to arrange in the middle.

3. **List the remaining letters:**
   The remaining letters after fixing P and S are: E, R, M, U, T, A, T, I, O, N. Notice that the letter 'T' still appears twice among these 10 letters.

4. **Calculate the number of distinct permutations:**
   The number of distinct permutations of 10 letters where 'T' is repeated twice is given by the formula:
   \[
   \frac{10!}{2!}
   \]
   Here, \(10!\) (10 factorial) is the total number of permutations of 10 letters, and \(2!\) (2 factorial) accounts for the indistinguishable arrangements caused by the two 'T's.

5. **Compute the factorials:**
   \[
   10! = 3,628,800
   \]
   \[
   2! = 2
   \]

6. **Divide to find the number of distinct permutations:**
   \[
   \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400
   \]

Therefore, the number of ways to arrange the letters of the word "PERMUTATIONS" such that the arrangement starts with P and ends with S is \(\boxed{1814400}\).

System: You are a helpful assistant.