Found problems: 49
2023 European Mathematical Cup, 4
We say that a $2023$-tuple of nonnegative integers $(a_1,\hdots,a_{2023})$ is [i]sweet[/i] if the following conditions hold:
[list]
[*] $a_1+\hdots+a_{2023}=2023$
[*] $\frac{a_1}{2}+\frac{a_2}{2^2}+\hdots+\frac{a_{2023}}{2^{2023}}\le 1$
[/list]
Determine the greatest positive integer $L$ so that \[a_1+2a_2+\hdots+2023a_{2023}\ge L\] holds for every sweet $2023$-tuple $(a_1,\hdots,a_{2023})$
[i]Ivan Novak[/i]
2023 LMT Fall, 5B
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]
LMT Theme Rounds, 2023F 1A
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]
2023 LMT Fall, 3A
A rectangular tea bag $PART$ has a logo in its interior at the point $Y$ . The distances from $Y$ to $PT$ and $PA$ are $12$ and $9$ respectively, and triangles $\triangle PYT$ and $\triangle AYR$ have areas $84$ and $42$ respectively. Find the perimeter of pentagon $PARTY$.
[i]Proposed by Muztaba Syed[/i]
[hide=Solution]
[i]Solution[/i]. $\boxed{78}$
Using the area and the height in $\triangle PYT$, we see that $PT = 14$, and thus $AR = 14$, meaning the height from $Y$ to $AR$ is $6$. This means $PA = TR = 18$. By the Pythagorean Theorem $PY=\sqrt{12^2+9^2} = 15$ and $YT =\sqrt{12^2 +5^2} = 13$. Combining all of these gives us an answer of $18+14+18+13+15 = \boxed{78}$.
[/hide]
2023 LMT Fall, 4C
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]
LMT Theme Rounds, 2023F 2C
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]
LMT Speed Rounds, 21
If $a \diamondsuit b = \vert a - b \vert \cdot \vert b - a \vert$ then find the value of $1 \diamondsuit (2 \diamondsuit (3 \diamondsuit (4 \diamondsuit 5)))$.
[i]Proposed by Muztaba Syed[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{9}$
$a\diamondsuit b = (a-b)^2$. This gives us an answer of $\boxed{9}$.
[/hide]
LMT Theme Rounds, 2023F 5B
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]
LMT Theme Rounds, 2023F 1C
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]
LMT Theme Rounds, 2023F 4C
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]
2023 LMT Fall, 1B
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]
2023 LMT Fall, 1A
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]
2023 LMT Fall, 2A
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]
2023 LMT Fall, 4
The numbers $1$, $2$, $3$, and $4$ are randomly arranged in a $2$ by $2$ grid with one number in each cell. Find the probability the sum of two numbers in the top row of the grid is even.
[i]Proposed by Muztaba Syed and Derek Zhao[/i]
[hide=Solution]
[i]Solution. [/i]$\boxed{\dfrac{1}{3}}$
Pick a number for the top-left. There is one number that makes the sum even no matter what we pick. Therefore, the
answer is $\boxed{\dfrac{1}{3}}$.[/hide]
2023 LMT Fall, 2
Eddie has a study block that lasts $1$ hour. It takes Eddie $25$ minutes to do his homework and $5$ minutes to play a game of Clash Royale. He can’t do both at the same time. How many games can he play in this study block while still completing his homework?
[i]Proposed by Edwin Zhao[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{7}$
Study block lasts 60 minutes, thus he has 35 minutes to play Clash Royale, during which he can play $\frac{35}{5}=\boxed{7}$ games.
[/hide]
2023 European Mathematical Cup, 1
Determine all sets of real numbers $S$ such that:
[list]
[*] $1$ is the smallest element of $S$,
[*] for all $x,y\in S$ such that $x>y$, $\sqrt{x^2-y^2}\in S$
[/list]
[i]Adian Anibal Santos Sepcic[/i]
2023 LMT Fall, 5C
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]
2023 LMT Fall, 5A
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]
2023 LMT Fall, 4A
Let [i]Revolution[/i]$(x) = x^3 +Ux^2 +Sx + A$, where $U$, $S$, and $A$ are all integers and $U +S + A +1 = 1773$. Given that [i]Revolution[/i] has exactly two distinct nonzero integer roots $G$ and $B$, find the minimum value of $|GB|$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{392}$
Notice that $U + S + A + 1$ is just [i]Revolution[/i]$(1)$ so [i]Revolution[/i]$(1) = 1773$. Since $G$ and $B$ are integer roots we write [i]Revolution[/i]$(X) = (X-G)^2(X-B)$ without loss of generality. So Revolution$(1) = (1-G)^2(1-B) = 1773$. $1773$ can be factored as $32 \cdot 197$, so to minimize $|GB|$ we set $1-G = 3$ and $1-B = 197$. We get that $G = -2$ and $B = -196$ so $|GB| = \boxed{392}$.
[/hide]
LMT Theme Rounds, 2023F 5A
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]
2023 European Mathematical Cup, 3
Let $n$ be a positive integer. Let $B_n$ be the set of all binary strings of length $n$. For a binary string $s_1\hdots s_n$, we define it's twist in the following way. First, we count how many blocks of consecutive digits it has. Denote this number by $b$. Then, we replace $s_b$ with $1-s_b$. A string $a$ is said to be a [i]descendant[/i] of $b$ if $a$ can be obtained from $b$ through a finite number of twists. A subset of $B_n$ is called [i]divided[/i] if no two of its members have a common descendant. Find the largest possible cardinality of a divided subset of $B_n$.
[i]Remark.[/i] Here is an example of a twist: $101100 \rightarrow 101000$ because $1\mid 0\mid 11\mid 00$ has $4$ blocks of consecutive digits.
[i]Viktor Simjanoski[/i]
LMT Speed Rounds, 3
Sam Wang decides to evaluate an expression of the form $x +2 \cdot 2+ y$. However, he unfortunately reads each ’plus’ as a ’times’ and reads each ’times’ as a ’plus’. Surprisingly, he still gets the problem correct. Find $x + y$.
[i]Proposed by Edwin Zhao[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{4}$
We have $x+2*2+y=x \cdot 2+2 \cdot y$. When simplifying, we have $x+y+4=2x+2y$, and $x+y=4$.
[/hide]
2023 LMT Fall, 3C
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]
LMT Speed Rounds, 2
Eddie has a study block that lasts $1$ hour. It takes Eddie $25$ minutes to do his homework and $5$ minutes to play a game of Clash Royale. He can’t do both at the same time. How many games can he play in this study block while still completing his homework?
[i]Proposed by Edwin Zhao[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{7}$
Study block lasts 60 minutes, thus he has 35 minutes to play Clash Royale, during which he can play $\frac{35}{5}=\boxed{7}$ games.
[/hide]
2023 LMT Fall, 2C
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]