BuilderBoi

Note the remainders for each term can be written in separate "cycles": 8, 7, 18, 11, 12, 1 

There are 336 complete "cycles" (note there are 2021 terms), each with a sum of 57 The last cycle has 5 terms, meaning it has a sum of 56. 

This means that the sum of their remainders is \(57 \times 336 + 56 = 19208\), meaning the remainder is \(19208 (\text{mod} \ 19) = \color{brown}\boxed{18}\)

LOL, I saw this in a MaxWong answer, and I've used it ever since

How many ways can we choose 3 members from the 13 people commitee?

How many ways can we choose 3 Democrats from the 6 Democrats? 

Note that when x = 1, y = a + b + c. 

Can you take it form here?

Yea, why stop at 1000? You need to catch up with Chris and Melody!!

It's been so fun seeing you grow!!!

Vieta's formula allows you to calculate the sum or product of the roots of the quadratic equation. 

For example, let's say we have the quadratic \(x^2 + 6x + 8\) (in the form \(ax^2 + bx + c\)). By Vieta's, the sum of the roots is \({-b \over a} = {-6 \over 1} = -6\), and the products of the roots is \({c \over a} = {8 \over 1} = 8\)

Hope this helps!!

The probability of getting 2 red marbles is \({3 \over 3 + k} \times {2 \over 2 + k} \)

The probability of getting 2 white marbles is \({k \over k + 3} \times { k - 1 \over k + 2}\)

This means that we have the equation \(\left(\frac{3}{3+k} \times \frac{2}{2+k}\right)+\left(\frac{k}{k + 3}\times \frac{k-1}{k+2}\right)=0.5\)

Now, just solve for k...

We have this equation: \(\frac{x}4-\frac{y}5=1\)

Subtract \({x \over 4}\) from both sides: \(-{y \over 5} = 1 - {x \over 4}\)

Multiply the entire equation by -5: \(y = {5 \over 4} x - 5\)

This means that the slope is \({5 \over 4}\), the slope of the line that is perpendicular to this is the negative reciprocal, or \(\color{brown}\boxed{-{4 \over 5}}\)

We have the system: 

\(2x + 5y = 11 \ \ \ \ \ \ \ (i)\)

\(3y = x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)\)

Start by substituting \((ii)\) into \((i)\), which gives us \(2(3y) + 5y = 11\), meaning \(y = 1\)

Now, plug in \(y = 1\) into \((ii)\), to get \(3 \times 1 = x\), meaning \(x = 3\).

This means that the solution is \((3,1) \), meaning the sum is \(3 + 1 = \color{brown}\boxed{4}\)