[hide=C stands for Cantor, G stands for Gauss]they had two problem sets under those two names[/hide] 


[b]C1.[/b] Mr. Pham flips $2018$ coins. What is the difference between the maximum and minimum number of heads that can appear?


[b]C2 / G1.[/b] Brandon wants to maximize $\frac{\Box}{\Box} +\Box$  by placing the numbers $1$, $2$, and $3$ in the boxes. If each number may only be used once, what is the maximum value attainable?


[b]C3.[/b] Guang has $10$ cents consisting of pennies, nickels, and dimes. What are all the possible numbers of pennies he could have?


[b]C4.[/b] The ninth edition of Campbell Biology has $1464$ pages. If Chris reads from the beginning of page $426$ to the end of page$449$, what fraction of the book has he read?


[b]C5 / G2.[/b] The planet Vriky is a sphere with radius $50$ meters. Kyerk starts at the North Pole, walks straight along the surface of the sphere towards the equator, runs one full circle around the equator, and returns to the North Pole. How many meters did Kyerk travel in total throughout his journey?


[b]C6 / G3.[/b] Mr. Pham is lazy and decides Stan’s quarter grade by randomly choosing an integer from $0$ to $100$ inclusive. However, according to school policy, if the quarter grade is less than or equal to $50$, then it is bumped up to $50$. What is the probability that Stan’s final quarter grade is $50$?


[b]C7 / G5.[/b] What is the maximum (finite) number of points of intersection between the boundaries of a equilateral triangle of side length $1$ and a square of side length $20$?


[b]C8.[/b] You enter the MBMT lottery, where contestants select three different integers from $1$ to $5$ (inclusive). The lottery randomly selects two winning numbers, and tickets that contain both of the winning numbers win. What is the probability that your ticket will win?


[b]C9 / G7.[/b] Find a possible solution $(B, E, T)$ to the equation $THE + MBMT = 2018$, where $T, H, E, M, B$ represent distinct digits from $0$ to $9$.


[b]C10.[/b] $ABCD$ is a unit square. Let $E$ be the midpoint of $AB$ and $F$ be the midpoint of $AD$. $DE$ and $CF$ meet at $G$. Find the area of $\vartriangle EFG$.


[b]C11.[/b] The eight numbers $2015$, $2016$, $2017$, $2018$, $2019$, $2020$, $2021$, and $2022$ are split into four groups of two such that the two numbers in each pair differ by a power of $2$. In how many different ways can this be done?


[b]C12 / G4.[/b] We define a function f such that for all integers $n, k, x$, we have that $$f(n, kx) = k^n f(n, x) and f(n + 1, x) = xf(n, x).$$ If $f(1, k) = 2k$ for all integers $k$, then what is $f(3, 7)$?


[b]C13 / G8.[/b] A sequence of positive integers is constructed such that each term is greater than the previous term, no term is a multiple of another term, and no digit is repeated in the entire sequence. An example of such a sequence would be $4$, $79$, $1035$. How long is the longest possible sequence that satisfies these rules?


[b]C14 / G11.[/b] $ABC$ is an equilateral triangle of side length $8$. $P$ is a point on side AB. If $AC +CP = 5 \cdot AP$, find $AP$.


[b]C15.[/b] What is the value of $(1) + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + ... + 49 + 50)$?


[b]G6.[/b] An ant is on a coordinate plane. It starts at $(0, 0)$ and takes one step each second in the North, South, East, or West direction. After $5$ steps, what is the probability that the ant is at the point $(2, 1)$?


[b]G10.[/b] Find the set of real numbers $S$ so that $$\prod_{c\in S}(x^2 + cxy + y^2) = (x^2 - y^2)(x^{12} - y^{12}).$$


[b]G12.[/b] Given a function $f(x)$ such that $f(a + b) = f(a) + f(b) + 2ab$ and $f(3) = 0$, find $f\left( \frac12 \right)$.


[b]G13.[/b] Badville is a city on the infinite Cartesian plane. It has $24$ roads emanating from the origin, with an angle of $15$ degrees between each road. It also has beltways, which are circles centered at the origin with any integer radius. There are no other roads in Badville. Steven wants to get from $(10, 0)$ to $(3, 3)$. What is the minimum distance he can take, only going on roads?


[b]G14.[/b] Team $A$ and Team $B$ are playing basketball. Team A starts with the ball, and the ball alternates between the two teams. When a team has the ball, they have a $50\%$ chance of scoring $1$ point. Regardless of whether or not they score, the ball is given to the other team after they attempt to score. What is the probability that Team $A$ will score $5$ points before Team $B$ scores any?


[b]G15.[/b] The twelve-digit integer $$\overline{A58B3602C91D},$$ where $A, B, C, D$ are digits with $A > 0$, is divisible by $10101$. Find $\overline{ABCD}$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the equation \( THE + MBMT = 2018 \) where \( T, H, E, M, B \) represent distinct digits from \( 0 \) to \( 9 \), we need to find a combination of digits that satisfies the equation. 

1. **Express the equation in terms of digits:**
   \[
   100T + 10H + E + 1000M + 100B + 10M + T = 2018
   \]
   Simplifying, we get:
   \[
   1001M + 100B + 101T + 10H + E = 2018
   \]

2. **Identify possible values for \( M \):**
   Since \( 1001M \) must be less than or equal to 2018, the possible values for \( M \) are \( 0, 1, \) and \( 2 \). However, since \( M \) must be a distinct digit and \( 1001M \) should be a significant part of 2018, we start with \( M = 1 \):
   \[
   1001 \times 1 = 1001
   \]
   Subtracting from 2018:
   \[
   2018 - 1001 = 1017
   \]

3. **Identify possible values for \( B \):**
   Since \( 100B \) must be less than or equal to 1017, the possible values for \( B \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, \) and \( 9 \). We start with \( B = 0 \):
   \[
   100 \times 0 = 0
   \]
   Subtracting from 1017:
   \[
   1017 - 0 = 1017
   \]

4. **Identify possible values for \( T \):**
   Since \( 101T \) must be less than or equal to 1017, the possible values for \( T \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, \) and \( 9 \). We start with \( T = 6 \):
   \[
   101 \times 6 = 606
   \]
   Subtracting from 1017:
   \[
   1017 - 606 = 411
   \]

5. **Identify possible values for \( H \):**
   Since \( 10H \) must be less than or equal to 411, the possible values for \( H \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, \) and \( 9 \). We start with \( H = 4 \):
   \[
   10 \times 4 = 40
   \]
   Subtracting from 411:
   \[
   411 - 40 = 371
   \]

6. **Identify possible values for \( E \):**
   Since \( E \) must be less than or equal to 371, the possible values for \( E \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, \) and \( 9 \). We start with \( E = 2 \):
   \[
   2 = 2
   \]
   Subtracting from 371:
   \[
   371 - 2 = 369
   \]

7. **Verify the solution:**
   \[
   100T + 10H + E = 100 \times 6 + 10 \times 4 + 2 = 600 + 40 + 2 = 642
   \]
   \[
   1000M + 100B + 10M + T = 1000 \times 1 + 100 \times 0 + 10 \times 1 + 6 = 1000 + 0 + 10 + 6 = 1016
   \]
   \[
   642 + 1016 = 1658 \neq 2018
   \]

8. **Adjust values and recheck:**
   After several trials, we find that:
   \[
   (T, H, E, M, B) = (6, 0, 2, 1, 4)
   \]
   works because:
   \[
   100 \times 6 + 10 \times 0 + 2 = 602
   \]
   \[
   1000 \times 1 + 100 \times 4 + 10 \times 1 + 6 = 1000 + 400 + 10 + 6 = 1416
   \]
   \[
   1416 + 602 = 2018
   \]

Thus, a possible solution is:
\[
(B, E, T) = \boxed{(4, 2, 6)}
\]