Danny likes seven-digit numbers with the following property: the 1's digit is divisible by the 10's digit, the 10's digit is divisible by the 100's digit, and so on. For example, Danny likes the number $1133366$ but doesn't like $9999993$. Is the amount of numbers Danny likes divisible by $7$?

There are $2017$ points on the plane, no three of them are collinear. Some pairs of the points are connected by $n$ segments. Find the smallest value of $n$ so that there always exists two disjoint segments in any case.

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Let $a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). \] Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

Bisector of one angle of triangle $ABC$ is equal to the bisector of its external angle at the same vertex (see figure). Find the difference between the other two angles of the triangle. [img]https://cdn.artofproblemsolving.com/attachments/c/3/d2efeb65544c45a15acccab8db05c8314eb5f2.png[/img]

Show that for any integter $a$, the number $\frac{a^5}5+\frac{a^3}3+\frac{7a}{15}$ is an integer.

Exactly $102$ country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least \(\frac{2}{3}\) of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly $81$ leaders present. The chairperson must seat them in a square-shaped conference hall of size \(9 \times 9\), where each leader will be seated in a designated \(1 \times 1\) cell. It is known that exactly $28$ of these $81$ leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells \(1 \times 1\) (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly $2$ neighbors and will vote "FOR" only if both of their neighbors voted "FOR." (a) Can the IMO chairperson arrange their $28$ supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least \(\frac{2}{3}\) of all $102$ leaders? (b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately? [i]Proposed by Bogdan Rublov[/i]

Let $ a,b,c,d$ be nonnegative real numbers. Prove that \[ a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}2abcd \ge a^2b^2\plus{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\plus{}c^2d^2.\]

Let be the polynomial $ f=X^4+X^2\in\mathbb{Z}_2[X] $ Find: a) its degree.. b) the splitting field of $ f $ c) the Galois group of $ f $ (Galois group of its splitting field)

[asy] size(5cm); defaultpen(fontsize(6pt)); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw((0,0)--(-4,0)--(-4,-4)--(0,-4)--cycle); draw((1,-1)--(1,3)--(-3,3)--(-3,-1)--cycle); draw((-1,1)--(-1,-3)--(3,-3)--(3,1)--cycle); draw((-4,-4)--(0,-4)--(0,-3)--(3,-3)--(3,0)--(4,0)--(4,4)--(0,4)--(0,3)--(-3,3)--(-3,0)--(-4,0)--cycle, red+1.2); label("1", (-3.5,0), S); label("2", (-2,0), S); label("1", (-0.5,0), S); label("1", (3.5,0), S); label("2", (2,0), S); label("1", (0.5,0), S); label("1", (0,3.5), E); label("2", (0,2), E); label("1", (0,0.5), E); label("1", (0,-3.5), E); label("2", (0,-2), E); label("1", (0,-0.5), E); [/asy] Compute the area of the resulting shape, drawn in red above. [i]Proposed by Nathan Xiong[/i]

What is the largest area of a regular hexagon that can be drawn inside the equilateral triangle of side $3$? A. $3\sqrt7$ B. $\frac{3 \sqrt3}{2}$ C. $2\sqrt5$ D. $\frac{3\sqrt3}{8}$ E. $3\sqrt5$

Given is a triangle $ABC$ with $AB>AC$. Its incircle touches $AB, AC$ at $D, E$, respectively. Let $CD$ meet the incircle at $K$ and $L$ is the foot of the perpendicular from $A$ to $CK$. If $M$ is the midpoint of $DE$ and $H$ is the orthocenter of $\triangle KML$, prove that $\angle AHK=90^{o}$. [i]Proposed by Dominik Burek[/i]

Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product? $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them $S$- and $Z$-tetrominoes, respectively. Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove that no matter how we tile $P$ using only $S$- and $Z$-tetrominoes, we always use an even number of $Z$-tetrominoes. [i]Proposed by Tamas Fleiner and Peter Pal Pach, Hungary[/i]

Let $A_1, A_2, ..., A_k$ be $5$-element subsets of set $\{1, 2, ..., 23\}$ such that, for all $1 \le i < j \le k$ set $A_i \cap A_j$ has at most three elements. Show that $k \le 2018$.