BuilderBoi

Let the starting term be s and the common difference be d. 

Then $a_3 = s + 2d$ and $a_5 = s + 4d$. 

We know that $3s + 6d = -12$, meaning that $s + 2d = -4 = a_3$

This means that $a_1 + a_5 = -8$ and their product is $80 \div -4 = -20$. 

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So, we have the system $a_1 + a_5 = -8$ and $a_1 \times a_5 = -20$. 

Solving these gives us $(a_1, a_5) = (-10, 2), (2, -10)$. 

In the first case, the starting term is -10 and the common difference is 3, so the 10th term is $-10 + 3 \times 9 = \color{brown}\boxed{17}$. 

In the second case, the starting term is 2 and the common difference is -3, so the 10th term is $2 - (3 \times 9) = \color{brown}\boxed{-25}$

My answer got deleted, so here we go again: (Note: I'm not too confident with this answer)

We find the number of cases where a pair of products is not a multiple of 3 or even and subtract that from the total number of cases (10!)

The only way for a pair of numbers to have a product that isn't even or a multiple of 3 is if two adjacent numbers are both of the set: {1 ,5, 7}

Split this up into 3 cases: 

Case 1 - All 3 numbers are next to each other: There are 3! ways to order them, 8 spots for this "block" and 7! ways to order the other numbers. This makes for $7! \times 8 \times 3! = 241920$ cases

Case 2 - A pair of 2 numbers are next to each other and are on the edge (spots 9 to 10 and 1 to 2): There are 3 ways to choose the numbers, 2 ways to order them, 7 spots for the third number, 2 spots to put the numbers, and 7! ways to order the rest. This makes for $7! \times 2 \times 7 \times 3 \times 2 = 423360$ cases.

Case 3 - A pair of 2 numbers are next to each other and they aren't on the edge: There are 3 ways to choose the numbers, 2 ways to order them, 7 spots to put them, 6 spots for the third number, and 7! ways to order the rest, which makes for $7! \times 6 \times 7 \times 3 \times 2 = 1270080$ cases.

So, there are $10! - 1270080 - 423360 - 241920 = \color{brown}\boxed{1,693,440}$cases.

Note that the only way to have a real number is to have both x and y have the same parity (look at the powers of i under guest's answer)

If both x and y are even, there are 50 odd numbers to choose from. Note that if we choose any pair of 2 distinct even numbers, there will be 1 case for x and y. So, in this case, there are ${50 \choose 2} = 1225$ cases. 

However, if x and y are odd, we need one to have a remainder of 1 when divided by 4 and the other to have a remainder of 3. There are 25 numbers for x and y, which makes for $25^2 = 625$ cases.

So, there are a total of $1225 + 625 = \color{brown}\boxed{1850}$ cases.

Divide it up by casework: 

If the twins get 0 pieces - There are 10 pieces of candy and 3 children for 10 stars and 2 bars which makes ${12 \choose 2} = 66$ cases

If the twins get 1 piece - There are 8 pieces of candy and 3 children for 8 stars and 2 bars which makes ${10 \choose 2} = 45$ cases

If the twins get 2 pieces - There are 6 pieces of candy and 3 children for 6 stars and 2 bars which makes ${8 \choose 2} = 28$ cases

If the twins get 3 pieces - There are 4 pieces of candy and 3 children for 4 stars and 2 bars which makes ${6 \choose 2} = 15$ cases

If the twins get 4 pieces - There are 2 pieces of candy and 3 children for 2 stars and 2 bars which makes ${4 \choose 2} = 6$ cases

If the twins get 5 pieces - There are 0 pieces of candy and 3 children for 0 stars and 2 bars which makes ${2 \choose 2} = 1$ case

So, there are $66 + 45 + 28 + 15 + 6 + 1 = \color{brown}\boxed{161}$ cases.

A diagram would help!

c) Note that each digit, except for 0, occurs 20 times - 10 times as a tens place and 10 times as a one place

So, the sum is just $20(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = \color{brown}\boxed{900}$

b) The only digits that are 0 occur in 10, 20, 30, ... , 90

There are 9 of these, so the probability is ${9 \over 189} = \color{brown}\boxed{1 \over 21}$

a)

Sam wrote 9 x 1 = 9 digits from 1 - 9

Sam wrote 90 x 2 = 180 digits from 10 - 99

So, he wrote a total of $180 + 9 = \color{brown}\boxed{189}$ digits

Because a side is perpendicular to a median, the triangle is isosceles. 

Because $PR = 14$, and $QN = 12$, the area of the triangle is $12 \times 14 \div 2 = \color{brown}\boxed{84}$

Note that PR is a base and QN is the height.

Being the 50th best means that there are 49 people better than Misha. 

Being the 50th worst means that there are 49 people worse than Misha. 

So, with Misha, there are $49 + 49 + 1=\color{brown}\boxed{99}$ students