Show that for any $n\geq 2$, $2^{2^n}+1$ ends with 7

In every cell of a table with $n$ rows and $m$ columns is written one of the letters $a$, $b$, $c$. Every two rows of the table have the same letter in at most $k\geq 0$ positions and every two columns coincide at most $k$ positions. Find $m$, $n$, $k$ if \[\frac{2mn+6k}{3(m+n)}\geq k+1\]

Each of $999$ numbers placed in a circular way is either $1$ or $-1$. (Both values appear). Consider the total sum of the products of every $10$ consecutive numbers. $(a)$ Find the minimal possible value of this sum. $(b)$ Find the maximal possible value of this sum.

Using only once each of the digits $1, 2, 3, 4, 5, 6, 7$ and $ 8$, write the square and the cube of a positive integer. Determine what that number can be.

Let $N$ be the number of sequences of natural numbers $d_1,d_2,\dots,d_{10}$ such that the following conditions hold: $d_1|d_2$, $\dots$, $d_9|d_{10}$ and $d_{10}|6^{2018}$. Evaluate the remainder when $N$ is divided by $2017$.

The absolute value of the difference of the solutions of the equation $x^2 + px + q = 0$, with $p, q \in R$, is equal to $4$. Find the solutions of the equation if it is known that $(q + 1) p^2 + q^2$ takes the minimum value.

Let $k$ be a given positive integer. Define $x_{1}= 1$ and, for each $n > 1$, set $x_{n+1}$ to be the smallest positive integer not belonging to the set $\{x_{i}, x_{i}+ik | i = 1, . . . , n\}$. Prove that there is a real number $a$ such that $x_{n}= [an]$ for all $n \in\mathbb{ N}$.

Triangle $ABC$ has $AB=10$, $BC=17$, and $CA=21$. Point $P$ lies on the circle with diameter $AB$. What is the greatest possible area of $APC$?

The ratio $ \dfrac{10^{2000}\plus{}10^{2002}}{10^{2001}\plus{}10^{2001}}$ is closest to which of the following numbers? $ \text{(A)}\ 0.1\qquad \text{(B)}\ 0.2\qquad \text{(C)}\ 1\qquad \text{(D)}\ 5\qquad \text{(E)}\ 10$

Let $ABC$ be a triangle with $AB = 2021, AC = 2022,$ and $BC = 2023.$ Compute the minimum value of $AP +2BP +3CP$ over all points $P$ in the plane.

The positive real numbers $a, b, c, d$ satisfy the equality $$\left(\frac{1}{a}+ \frac{1}{b}\right) \left(\frac{1}{c}+ \frac{1}{d}\right) + \frac{1}{ab}+ \frac{1}{cd} = \frac{6}{\sqrt{abcd}}$$ Find the value of the $$\frac{a^2+ac+c^2}{b^2-bd+d^2}$$

Triangle $ ABC$ and point $ P$ in the same plane are given. Point $ P$ is equidistant from $ A$ and $ B$, angle $ APB$ is twice angle $ ACB$, and $ \overline{AC}$ intersects $ \overline{BP}$ at point $ D$. If $ PB \equal{} 3$ and $ PD \equal{} 2$, then $ AD\cdot CD \equal{}$ $ \textbf{(A)}\ 5\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$ [asy]defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]

Solve: $x^2(2-x)^2=1+2(1-x)^2$ Where $x$ is real number.

Let $ABC$ be an acute triangle with $AB< AC$. Denote by $P$ and $Q$ points on the segment $BC$ such that $\angle BAP = \angle CAQ < \frac{\angle BAC}{2}$. $B_1$ is a point on segment $AC$. $BB_1$ intersects $AP$ and $AQ$ at $P_1$ and $Q_1$, respectively. The angle bisectors of $\angle BAC$ and $\angle CBB_1$ intersect at $M$. If $PQ_1\perp AC$ and $QP_1\perp AB$, prove that $AQ_1MPB$ is cyclic.

Two circles $\omega_1$ and $\omega_2$, which have centers $O_1$ and $O_2$, respectively, intersect at $A$ and $B$. A line $\ell$ that passes through $B$ cuts to $\omega_1$ again at $C$ and cuts to $\omega_2$ again in $D$, so that points $C, B, D$ appear in that order. The tangents of $\omega_1$ and $\omega_2$ in $C$ and $D$, respectively, intersect in $E$. Line $AE$ intersects again to the circumscribed circumference of the triangle $AO_1O_2$ in $F$. Try that the length of the $EF$ segment is constant, that is, it does not depend on the choice of $\ell$.

Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$.