BuilderBoi

Write as system: 

x = cost of first hotel

y = cost of second hotel

0.075x + 0.095y = 542.25

y = x - 1000

SOlving, we find $\color{brown}\boxed{x=3,750, y = 2,750}$

I think it is $11 \choose 7$. There are 7 ways to choose the boys with orange caps, and the rest have yellow caps. 

Thus, there is $\color{brown}\boxed{330}$ ways. (Note, ${11 \choose 7} = {11 \choose 4}$, so the answer would be the same either way. 

Here's a better way to get the same answer. 

For groups of 2 including  ₯, there is $4 \choose 1$ choices, because 1 of them is already taken. 

For groups of 3 including  ₯, there is $4 \choose 2$ choices, because 1 of them is already taken. 

For groups of 4 including  ₯, there is $4 \choose 3$ choices, because 1 of them is already taken. 

For groups of 5 including  $₯$, there is $4 \choose 4$ choices, because 1 of them is already taken. 

Adding everything up, there is $\color{brown}\boxed{15}$ subsets that contain ₯.

I will number these from right to left, starting with 1, going up to 5. The rightmost one is 1, the leftmost is 5. 

2 1

2 3

2 4

2 5 

2 1 3 

2 1 4 

2 1 5 

2 3 4 

2 3 5 

2 4 5

2 1 3 4 

2 1 3 5 

2 1 4 5 

2 3 4 5 

2 1 3 4 5 

Thus, there is a total of $\color{brown}\boxed{15}$ subsets with the ₯ symbol. 

It has to be in the form ABCBA. There are 5 choices for A, 5 for B, and 5 for C, making a total of $\color{brown}\boxed{125}$ choices.

There are 2 cases: The second digit is a 2 or it is a 4

If the second digit is a 2: $9 \times 1 \times 10 \times 5 = 450$

If the second digit is a 4: $9 \times 1 \times 10 \times 9 = 810$

Add these together for a total of $\color{brown}\boxed{1,260}$ ways

HAPPY ANIVERSARY CPHILL!!!

You made it so easy to learn new math skills and topics!!!

Because it's a right triangle, and the remaining angles are equal, this is a 45-45-90 triangle, with a width and heigth of 6, making the area $6 \times 6 \div 2 = \color{brown}\boxed{18}$

Write as an equation: 

${2 \over 5} ={ x + 1000 \over 2x+ 1000}$

Solving for x, we find John originally had $\color{brown}\boxed{-3,000}$ dollars. 

Here's another way to prove it's impossible:

$k \times 1984 \times 2 = k \times 2^7 \times 31$

Using this logic, you can create the factors: 1, 2, 4, 8, 16, 31, 32, 62, 64, 124, 128, 248, 496, 992, 1984, and 3968.

Because we want 21 factors, we need the number to be a perfect square. The smallest way to do this is by multiplying $3968 \times 31 \times 2$. This yields 27 factors, way too big. Any bigger and you get more factors. 

Thus, this problem is impossible, like geno said (Note: If this problem was just $k \times 1984$, we could multiply by 31, and get 21 factors. )