Let $a_1<a_2<a_3<\dots$ be positive integers such that $a_{k+1}$ divides $2(a_1+a_2+\dots+a_k)$ for every $k\geqslant 1$. Suppose that for infinitely many primes $p$, there exists $k$ such that $p$ divides $a_k$. Prove that for every positive integer $n$, there exists $k$ such that $n$ divides $a_k$.

For a positive integer $n$, let $P(n)$ be the product of the factors of $n$ (including $n$ itself). A positive integer $n$ is called [i]deplorable [/i] if $n > 1$ and $\log_n P(n)$ is an odd integer. How many factors of $2016$ are [i]deplorable[/i]?

Let $n\geqslant 2$ be an integer and $A{}$ a set of $n$ points in the plane. Find all integers $1\leqslant k\leqslant n-1$ with the following property: any two circles $C_1$ and $C_2$ in the plane such that $A\cap\text{Int}(C_1)\neq A\cap\text{Int}(C_2)$ and $|A\cap\text{Int}(C_1)|=|A\cap\text{Int}(C_2)|=k$ have at least one common point. [i]Cristi Săvescu[/i]

Logan is constructing a scaled model of his town. The city's water tower stands $ 40$ meters high, and the top portion is a sphere that holds $ 100,000$ liters of water. Logan's miniature water tower holds $ 0.1$ liters. How tall, in meters, should Logan make his tower? $ \textbf{(A)}\ 0.04\qquad \textbf{(B)}\ \frac{0.4}{\pi}\qquad \textbf{(C)}\ 0.4\qquad \textbf{(D)}\ \frac{4}{\pi}\qquad \textbf{(E)}\ 4$

Find $$\sum^{k=672}_{k=0} { 2018\choose {3k+2}} \,\, (mod \, 3)$$

Let $p,q,r$ be distinct prime numbers and let \[A=\{p^aq^br^c\mid 0\le a,b,c\le 5\} \] Find the least $n\in\mathbb{N}$ such that for any $B\subset A$ where $|B|=n$, has elements $x$ and $y$ such that $x$ divides $y$. [i]Ioan Tomescu[/i]

A square $ABCD$ is given. A point $P$ is chosen inside the triangle $ABC$ such that $\angle CAP = 15^\circ = \angle BCP$. A point $Q$ is chosen such that $APCQ$ is an isosceles trapezoid: $PC \parallel AQ$, and $AP=CQ, AP\nparallel CQ$. Denote by $N$ the midpoint of $PQ$. Find the angles of the triangle $CAN$.

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]

We are searching for the number $7$ in the following binary tree: [center] [img] https://cdn.artofproblemsolving.com/attachments/8/c/70ad159d239e9fd8dd9775e6391965e1016f03.png [/img] [/center] We use the following algorithm (which terminates with probability $1$): [list=1] [*] Write down the number currently at the root node. [*] If we wrote down $7,$ terminate. [*] Else, pick a random edge, and swap the two numbers at the endpoints of that edge [*] Go back to step $1.$ [/list] Let $p(a)$ be the probability that we ever write down the number $a$ after running the algorithm once. Find $$p(1)+p(2)+p(3)+p(5)+p(6).$$

Find the smallest positive integer $n$ such that there are at least three distinct ordered pairs $(x,y)$ of positive integers such that \[x^2-y^2=n.\]

If $$\Omega_n=\sum \limits_{k=1}^n \left(\int \limits_{-\frac{1}{k}}^{\frac{1}{k}}(2x^{10} + 3x^8 + 1)\cos^{-1}(kx)dx\right)$$Then find $$\Omega=\lim \limits_{n\to \infty}\left(\Omega_n-\pi H_n\right)$$

(F.Nilov) Given isosceles triangle $ ABC$ with base $ AC$ and $ \angle B \equal{} \alpha$. The arc $ AC$ constructed outside the triangle has angular measure equal to $ \beta$. Two lines passing through $ B$ divide the segment and the arc $ AC$ into three equal parts. Find the ratio $ \alpha / \beta$.

Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have \[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]

Let $\Delta ABC$ be a triangle with orthocenter $H$ and $\Gamma$ be the circumcircle of $\Delta ABC$ with center $O$. Consider $N$ the center of the circle that passes through the feet of the heights of $\Delta ABC$ and $P$ the intersection of the line $AN$ with the circle $\Gamma$. Suppose that the line $AP$ is perpendicular to the line $OH$. Prove that $P$ belongs to the reflection of the line $OH$ by the line $BC$.

A cow lives on a cubic planet of side length $12$. It is tied on a leash $12$ units long that is staked at the center of one of the faces of the cube. The total surface area that the cow can graze is $A \pi+B( \sqrt3 -1)$. Find $A + B$.

Jerry is mowing a rectangular lawn which is $77$ feet north to south by $83$ feet east to west. His lawn mower cuts a path $18$ inches wide. Jerry mows the grass by cutting a path from west to east across the north side of the lawn and then making a right turn cutting a path along the east side of the lawn. When he completes mowing each side of the lawn, he continues by making right turns to mow a path along the next side. How many right turns will he make?