A person starting with $64$ cents and making $6$ bets, wins three times and loses three times, the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: $\textbf{(A)}\ \text{a loss of } 27 \qquad \textbf{(B)}\ \text{a gain of }27 \qquad \textbf{(C)}\ \text{a loss of }37 \qquad$ $ \textbf{(D)}\ \text{neither a gain nor a loss} \qquad \textbf{(E)}\ \text{a gain or a loss depending upon the order in which the wins and losses occur}$ Note: Due to the lack of $\LaTeX$ packages, the numbers in the answer choices are in cents ¢

Let $n$ be a positive integer. - Find, in terms of $n$, the number of pairs $(x,y)$ of positive integers that are solutions of the equation : $$x^2-y^2=10^2.30^{2n}$$ - Prove further that this number is never a square

A $24$-hour digital clock shows times $h : m : s$, where $h$, $m$, and $s$ are integers with $0 \le h \le 23$, $0 \le m \le 59$, and $0 \le s \le 59$. How many times $h : m : s$ satisfy $h + m = s$?

Let $ABC$ be an acute scalene triangle. Let $D, E$ be points on segments $AB, AC$ respectively, such that $BD=CE$. Prove that the nine-point centers of $ADE$, $ACD$, $ABC$, $AEB$ form a rhombus. [i]Proposed by Malay Mahajan and Siddharth Choppara[/i]

The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$. [asy] import graph; size(4cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); draw((1.13,2.5)--(-0.13,2.5)); draw((-0.13,2.5)--(-0.13,1.23)); draw((-0.13,1.23)--(1.13,1.23)); draw((1.13,1.23)--(1.13,2.5)); [/asy]

Each square in an $n \times n$ table is painted black or white. The routes where two rows meet two columns, called a quartet if the remaining squares are the same color. (a) What is the largest possible number of black squares in a $4 \times 4$ table without quartets? (b) Is it possible to paint a $5 \times 5$ table so that it has no quartets?

Define a $\textit{subsequence}$ of a string $\mathcal{S}$ of letters to be a positive-lenght string using any number of the letters in $\mathcal{S}$ in order. For example, a subsequence of $HARRISON$ is $ARRON$. Compute the number of subsequences in $HARRISON$.

A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$? ${ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{\sqrt{6}}{2} \qquad\textbf{(C)}\ \sqrt{3}\qquad\textbf{(D}}\ 2 \qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} $

Let $I$ be the incenter of the scalene $\Delta ABC$, such, $AB<AC$, and let $I'$ be the reflection of point $I$ in line $BC$. The angle bisector $AI$ meets $BC$ at $D$ and circumcircle of $\Delta ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove, that, $\text{(i) } \frac{AI}{IE}=\frac{ID}{DE}$ $\text{(ii) } IA=IF$

$20.$ The circle $\omega$ touches the circle $\Omega$ internally at point $P.$ The centre $O$ of $\Omega$ is outside $\omega.$ Let $XY$ be a diameter of $\Omega$ which is also tangent to $\omega.$ Assume $PY>PX.$ Let $PY$ intersect $\omega$ at $z.$ If $YZ=2PZ,$ what is the magnitude of $\angle{PYX}$ in degrees $?$

Lazim rolls two $24$-sided dice. From the two rolls, Lazim selects the die with the highest number. $N$ is an integer not greater than $24$. What is the largest possible value for $N$ such that there is a more than $50$% chance that the die Lazim selects is larger than or equal to $N$?

Show that $n!=a^{n-1}+b^{n-1}+c^{n-1}$ has only finitely many solutions in positive integers. [i]Proposed by Dorlir Ahmeti, Albania[/i]

Let \( u_n \) be the \( n \)-th term of the Fibonacci sequence (where \( u_1 = u_2 = 1 \) and \( u_{n+1} = u_n + u_{n-1} \) for \( n \geq 2 \)). For each prime \( p \), let \( n(p) \) be the smallest integer \( n \) such that \( u_n \) is divisible by \( p \). Find the smallest possible value of \( p - n(p) \).

A unit cube has vertices $P_1, P_2, P_3, P_4, P_1', P_2', P_3'$, and $P_4'$. Vertices $P_2, P_3$, and $P_4$ are adjacent to $P_1$, and for $1\leq i\leq 4$, vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2, P_1P_3, P_1P_4, P_1'P_2', P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length? [asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label("$P_1$", P[1], dir(P[1])); label("$P_2$", P[2], dir(P[2])); label("$P_3$", P[3], dir(-45)); label("$P_4$", P[4], dir(P[4])); label("$P'_1$", Pp[1], dir(Pp[1])); label("$P'_2$", Pp[2], dir(Pp[2])); label("$P'_3$", Pp[3], dir(-100)); label("$P'_4$", Pp[4], dir(Pp[4])); [/asy] $ \textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2} $

$ABC$ is an isosceles triangle with $AB = AC$. Let $P$ be any point of a circle tangent to the sides $AB$ in $B$ and to AC in C. Denote $a$, $b$ and $c$ to the distances from $P$ to the sides $BC, AC$ and $AB$ respectively. Prove that: $a^2=bc$

Compute the sum of all twenty-one terms of the geometric series \[1+2+4+8+\cdots+1048576.\] $\textbf{(A) }2097149\hspace{12em}\textbf{(B) }2097151\hspace{12em}\textbf{(C) }2097153$ $\textbf{(D) }2097157\hspace{12em}\textbf{(E) }2097161$

Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots?

Given a positive integer $n$, determine how many permutations $\sigma$ of the set $\{1, 2, \ldots , 2022n\}$ have the following property: for each $i \in \{1, 2, \ldots , 2021n + 1\}$, the number $$\sigma(i) + \sigma(i + 1) + \cdots + \sigma(i + n - 1)$$ is a multiple of $n$.

Triangle $ABC$ is the right angled triangle with the vertex $C$ at the right angle. Let $P$ be the point of reflection of $C$ about $AB$. It is known that $P$ and two midpoints of two sides of $ABC$ lie on a line. Find the angles of the triangle.

a) Suppose that $ RBR'B'$ is a convex quadrilateral such that vertices $ R$ and $ R'$ have red color and vertices $ B$ and $ B'$ have blue color. We put $ k$ arbitrary points of colors blue and red in the quadrilateral such that no four of these $ k\plus{}4$ point (except probably $ RBR'B'$) lie one a circle. Prove that exactly one of the following cases occur? 1. There is a path from $ R$ to $ R'$ such that distance of every point on this path from one of red points is less than its distance from all blue points. 2. There is a path from $ B$ to $ B'$ such that distance of every point on this path from one of blue points is less than its distance from all red points. We call these two paths the blue path and the red path respectively. Let $ n$ be a natural number. Two people play the following game. At each step one player puts a point in quadrilateral satisfying the above conditions. First player only puts red point and second player only puts blue points. Game finishes when every player has put $ n$ points on the plane. First player's goal is to make a red path from $ R$ to $ R'$ and the second player's goal is to make a blue path from $ B$ to $ B'$. b) Prove that if $ RBR'B'$ is rectangle then for each $ n$ the second player wins. c) Try to specify the winner for other quadrilaterals.

Let $A_1C_2B_1A_2C_1B_2$ be an equilateral hexagon. Let $O_1$ and $H_1$ denote the circumcenter and orthocenter of $\triangle A_1B_1C_1$, and let $O_2$ and $H_2$ denote the circumcenter and orthocenter of $\triangle A_2B_2C_2$. Suppose that $O_1 \ne O_2$ and $H_1 \ne H_2$. Prove that the lines $O_1O_2$ and $H_1H_2$ are either parallel or coincide. [i]Ankan Bhattacharya[/i]

Given a positive integer $n$. One of the roots of a quadratic equation $x^{2}-ax +2 n = 0$ is equal to $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}$. Prove that $2\sqrt{2n}\le a\le 3\sqrt{n}$

Let \[x = .123456789101112\ldots998999,\] where the digits are obtained by writing the integers 1 through 999 in order. The 1983rd digit to the right of the decimal point is $ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 $

Find the unique real number $c$ such that the polynomial $x^3+cx+c$ has exactly two real roots.

Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime.