Found problems: 42
LMT Theme Rounds, 2023F 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]
LMT Theme Rounds, 2023F 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]
2023 LMT Fall, 3B
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]
LMT Speed Rounds, 6
Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{10}$
The chance of getting any given number is $\dfrac{1}{n}$
, so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]
LMT Theme Rounds, 2023F 4B
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]
LMT Theme Rounds, 2023F 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, 6
Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{10}$
The chance of getting any given number is $\dfrac{1}{n}$
, so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]
2023 LMT Fall, 1
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 Speed Rounds, 5
Let $a$ and $b$ be two-digit positive integers. Find the greatest possible value of $a+b$, given that the greatest common factor of $a$ and $b$ is $6$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{186}$
We can write our two numbers as $6x$ and $6y$. Notice that $x$ and $y$ must be relatively prime. Since $6x$ and $6y$ are two digit numbers, we just need to check values of $x$ and $y$ from $2$ through $16$ such that $x$ and $y$ are relatively prime. We maximize the sum when $x = 15$ and $y = 16$, since consecutive numbers are always relatively prime. So the sum is $6 \cdot (15+16) = \boxed{186}$.[/hide]
2023 LMT Fall, 2B
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]
2023 LMT Fall, 5
Let $a$ and $b$ be two-digit positive integers. Find the greatest possible value of $a+b$, given that the greatest common factor of $a$ and $b$ is $6$.
[i]Proposed by Jacob Xu[/i]
[hide=Solution][i]Solution[/i]. $\boxed{186}$
We can write our two numbers as $6x$ and $6y$. Notice that $x$ and $y$ must be relatively prime. Since $6x$ and $6y$ are two digit numbers, we just need to check values of $x$ and $y$ from $2$ through $16$ such that $x$ and $y$ are relatively prime. We maximize the sum when $x = 15$ and $y = 16$, since consecutive numbers are always relatively prime. So the sum is $6 \cdot (15+16) = \boxed{186}$.[/hide]
LMT Theme Rounds, 2023F 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]
2023 LMT Fall, 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 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]
LMT Theme Rounds, 2023F 3B
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]
LMT Theme Rounds, 2023F 2B
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]
LMT Theme Rounds, 2023F 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]