For a real number $x$, the minimum value of the expression $$\frac{2x^2 + x - 3}{x^2 - 2x + 3}$$ can be written in the form $\frac{a-\sqrt{b}}{c}$, where $a, b$, and $c$ are positive integers such that $a$ and $c$ are relatively prime. Find $a + b + c$

In Ace Attorney, Phoenix Wright is rolling a standard fair $20$-sided die. He can roll this die up to three times. After each roll, Phoenix can yell "Objection!'' to roll again, or "Hold It!'' to stop and keep his current number. If Phoenix plays optimally to maximize his final number, find the expected value of this number.

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

A four-digit number $n$ is said to be [i]literally 1434[/i] if, when every digit is replaced by its remainder when divided by $5$, the result is $1434$. For example, $1984$ is [i]literally 1434[/i] because $1$ mod $5$ is $1$, $9$ mod $5$ is $4$, $8$ mod $5$ is $3$, and $4$ mod $5$ is $4$. Find the sum of all four-digit positive integers that are [i]literally 1434[/i]. [i]Proposed by Evin Liang[/i] [hide=Solution] [i]Solution.[/i] $\boxed{67384}$ The possible numbers are $\overline{abcd}$ where $a$ is $1$ or $6$, $b$ is $4$ or $9$, $c$ is $3$ or $8$, and $d$ is $4$ or $9$. There are $16$ such numbers and the average is $\dfrac{8423}{2}$, so the total in this case is $\boxed{67384}$. [/hide]

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]

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]

Paul Revere is currently at $\left(x_0, y_0\right)$ in the Cartesian plane, which is inside a triangle-shaped ship with vertices at $\left(-\dfrac{7}{25},\dfrac{24}{25}\right),\left(-\dfrac{4}{5},\dfrac{3}{5}\right)$, and $\left(\dfrac{4}{5},-\dfrac{3}{5}\right)$. Revere has a tea crate in his hands, and there is a second tea crate at $(0,0)$. He must walk to a point on the boundary of the ship to dump the tea, then walk back to pick up the tea crate at the origin. He notices he can take 3 distinct paths to walk the shortest possible distance. Find the ordered pair $(x_0, y_0)$. [i]Proposed by Derek Zhao[/i] [hide=Solution][i]Solution.[/i] $\left(-\dfrac{7}{25},\dfrac{6}{25}\right)$ Let $L$, $M$, and $N$ be the midpoints of $BC$, $AC$, and $AB$, respectively. Let points $D$, $E$, and $F$ be the reflections of $O = (0,0)$ over $BC$, $AC$, and $AB$, respectively. Notice since $MN \parallel BC$, $BC \parallel EF$. Therefore, $O$ is the orthocenter of $DEF$. Notice that $(KMN)$ is the nine-point circle of $ABC$ because it passes through the midpoints and also the nine-point circle of $DEF$ because it passes through the midpoints of the segments connecting a vertex to the orthocenter. Since $O$ is both the circumcenter of $ABC$ and the orthocenter of $DEF$ and the triangles are $180^\circ$ rotations of each other, Revere is at the orthocenter of $ABC$. The answer results from adding the vectors $OA +OB +OC$, which gives the orthocenter of a triangle.[/hide]

Suppose $h$, $i$, $o$ are real numbers that satisfy the products $hi = 12$, $ooh = 18$, and $hohoho = 27$. Find the value of the product $ohio$.

How many distinct triangles are there with prime side lengths and perimeter $100$? [i]Proposed by Muztaba Syed[/i] [hide=Solution][i]Solution.[/i] $\boxed{0}$ As the perimeter is even, $1$ of the sides must be $2$. Thus, the other $2$ sides are congruent by Triangle Inequality. Thus, for the perimeter to be $100$, both of the other sides must be $49$, but as $49$ is obviously composite, the answer is thus $\boxed{0}$.[/hide]

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]

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]

The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$. [i]Proposed by Christopher Cheng[/i] [hide=Solution][i]Solution. [/i] $\boxed{6}$ Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide]

A four-digit number $n$ is said to be [i]literally 1434[/i] if, when every digit is replaced by its remainder when divided by $5$, the result is $1434$. For example, $1984$ is [i]literally 1434[/i] because $1$ mod $5$ is $1$, $9$ mod $5$ is $4$, $8$ mod $5$ is $3$, and $4$ mod $5$ is $4$. Find the sum of all four-digit positive integers that are [i]literally 1434[/i]. [i]Proposed by Evin Liang[/i] [hide=Solution] [i]Solution.[/i] $\boxed{67384}$ The possible numbers are $\overline{abcd}$ where $a$ is $1$ or $6$, $b$ is $4$ or $9$, $c$ is $3$ or $8$, and $d$ is $4$ or $9$. There are $16$ such numbers and the average is $\dfrac{8423}{2}$, so the total in this case is $\boxed{67384}$. [/hide]

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]

Consider the addition $\begin{tabular}{cccc} & O & N & E \\ + & T & W & O \\ \hline F & O & U & R \\ \end{tabular}$ where different letters represent different nonzero digits. What is the smallest possible value of the four-digit number $FOUR$?

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]

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 $\overline{SKIBIDI}$.

For a real number $x$, the minimum value of the expression $$\frac{2x^2 + x - 3}{x^2 - 2x + 3}$$ can be written in the form $\frac{a-\sqrt{b}}{c}$, where $a, b$, and $c$ are positive integers such that $a$ and $c$ are relatively prime. Find $a + b + c$

Kendrick Lamar and Drake are cutting their circular beef to share with their fans. The cuts must pass all the way from one side of the beef to the other, and no other modifications may be performed on the beef (e.g. folding, eating, stacking, etc.). Find the minimum number of cuts they will need to split their beef into $2024$ pieces.

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]

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]

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?

In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Let $M$ be the midpoint of side $AB$, $G$ be the centroid of $\triangle ABC$, and $E$ be the foot of the altitude from $A$ to $BC$. Compute the area of quadrilateral $GAME$. [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution[/i]. $\boxed{23}$ Use coordinates with $A = (0,12)$, $B = (5,0)$, and $C = (-9,0)$. Then $M = \left(\dfrac{5}{2},6\right)$ and $E = (0,0)$. By shoelace, the area of $GAME$ is $\boxed{23}$.[/hide]

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