The number $2020$ has three different prime factors. What is their sum?

Let $R$ be the rectangle on the cartesian plane with vertices $(0,0)$, $(5,0)$, $(5,7)$, and $(0,7)$. Find the number of squares with sides parallel to the axes and vertices that are lattice points that lie within the region bounded by $R$. [i]Proposed by Boyan Litchev[/i] [hide=Solution][i]Solution[/i]. $\boxed{85}$ We have $(6-n)(8-n)$ distinct squares with side length $n$, so the total number of squares is $5 \cdot 7+4 \cdot 6+3 \cdot 5+2 \cdot 4+1\cdot 3 = \boxed{85}$.[/hide]

Kanye West's favorite positive integer this year is $c$, and last year it was $c-t=20011$ (a prime), for some positive integer $t$ relatively prime to $c$. His two most streamed albums got $a$ and $b$ streams this year and $a-t$ and $b-t$ streams last year with $a > b > c$. Suppose $a \le 1.6 \times 10^9$ and his favorite integer in each year divides the number of streams for both albums in the corresponding year. Find the largest possible value of $c$.

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]

Let $S$, $K$, $I$, $B$, $D$, $Y$ be distinct integers from $0$ to $9,$ inclusive. Given that they follow this equation: $$\begin{array}{rrrrr} & S & K & I & B \\ - & I & D & I & D \\ \hline & & & D & Y \end{array}$$find the maximum value of $\overline{SKIBIDI}$.

In Brawl Stars, Rico can shoot his opponent directly or make his bullet bounce off the wall at the same angle. His opponent is $15$ feet in front of him and there are infinitely long walls $1$ feet to the left and right of Rico. If Rico's bullet travels $d$ feet before hitting the opponent, find the sum of all possible integer values of $d$.

Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$ [i]Proposed by Jonathan Liu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{64}$ We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$ [/hide]

The number $2020$ has three different prime factors. What is their sum?

Sam dumps tea for $6$ hours at a constant rate of $60$ tea crates per hour. Eddie takes $4$ hours to dump the same amount of tea at a different constant rate. How many tea crates does Eddie dump per hour? [i]Proposed by Samuel Tsui[/i] [hide=Solution] [i]Solution.[/i] $\boxed{90}$ Sam dumps a total of $6 \cdot 60 = 360$ tea crates and since it takes Eddie $4$ hours to dump that many he dumps at a rate of $\dfrac{360}{4}= \boxed{90}$ tea crates per hour. [/hide]

In Genshin Impact, $PRIMOGEM'$ is the octagon in the diagram below. Let $A$ be the intersection of $PO$ and $IE$. Suppose $PR=RI=IM=MO=OG=GE=EM'=M'P$, $AP=AI=AO=AE=4$, and $AR=AM=AG=AM'=\sqrt{2}$. Find the area of $PRIMOGEM'$. [asy] size(5cm); pair P = (0, 4), R = (1, 1), I = (4, 0), M = (1, -1), O = (0, -4), G = (-1, -1), E = (-4, 0), MM = (-1, 1), origin = (0, 0); draw(P--R--I--M--O--G--E--MM--P); draw(origin--P); draw(origin--I); draw(origin--O); draw(origin--E); draw(R--G); draw(MM--M); label("$P$", P, N); label("$R$", R, NE); label("$I$", I, SE); label("$M$", M, SE); label("$G$", G, SW); label("$E$", E, W); label("$M'$", MM, NW); label("$O$", O, S); [/asy]

Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost? [i]Proposed by Edwin Zhao[/i] [hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$ Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$. At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is $$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$ [/hide]

Travis Scott says "FEIN'' every $0.8$ seconds. Find the tens digit of the number of times he says "FEIN'' in $1$ minute.

Eminem is trying to find the real Slim Shady in a row of $2025$ indistinguishable Slim Shady clones, one of which is the real Slim Shady. Eminem randomly guesses, and if he guesses wrong, a new clone joins the row and all the clones randomly rearrange themselves. He keeps guessing as more identical clones are added, trying to find the real Slim Shady. Find the probability that he will eventually find him within $15$ guesses.

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]

The Den has two deals on chicken wings. The first deal is $4$ chicken wings for $3$ dollars, and the second deal is $11$ chicken wings for $ 8$ dollars. If Jeremy has $18$ dollars, what is the largest number of chicken wings he can buy?

Determine the least integer $n$ such that for any set of $n$ lines in the 2D plane, there exists either a subset of $1001$ lines that are all parallel, or a subset of $1001$ lines that are pairwise nonparallel. [i]Proposed by Samuel Wang[/i] [hide=Solution][i]Solution.[/i] $\boxed{1000001}$ Since being parallel is a transitive property, we note that in order for this to not exist, there must exist at most $1001$ groups of lines, all pairwise intersecting, with each group containing at most $1001$ lines. Thus, $n = 1000^2 + 1 = \boxed{1000001}$.[/hide]

On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons? [i]Proposed by Aidan Duncan[/i] [hide=Solution] [i]Solution. [/i] $\boxed{13}$ The total amount that John and Samuel have drank by day $n$ is $$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer. [/hide]