A sequence $\{a_n\}$ is defined as follows: $a_1 = 1$ and $a_{n+1} = 420a_n - 69$ for all positive integers $n$. The number of squares in this sequence is $k$. Enter $k$. (If there are infinitely many square numbers in the sequence, enter 999.)

Faces of a rectangular prism are coloured such that one face is red, two faces are green and three faces are blue. The area of the red face is $100$, the sum of areas of the green faces is $116$ and the sum of areas of the blue faces is $114$. It is known that the volume of the rectangular prism, $V$, is an integer. Enter $V$.

Let $\vartriangle ABC$ be a triangle. $I$ is the incentre of $\vartriangle ABC$, $BI$ and $AC$ meet at $D$, $CI$ and $AB$ meet at $E$. It is known that $DE = 12$ and $ADEI$ are concyclic. Suppose the circumradius of the $ADEI$ is $R$. Enter $R^2$.

Define $Alt(n) = d_1 - d_2 + d_3 - d_4 + ... + (-1)^{n+1}d_n$, where $d_1$, $d_2$, $...$, $d_n$ are the digits of n from left to right (for example if $n = 123$, then $d_1 = 1$, $d_2 = 2$, $d_3 = 3$). Now we define $a_k$ as the smallest natural number $m$ such that $Alt(m) = k$. Amongst $a_1, a_2, . . . a_{2023}$, $s$ of them are divisible by $11$. Enter $s$ mod $1000$.

Let $a_1, a2, . . . a_{2023}$ be $2023$ non-zero real numbers. Then, define $b_1, b_2, . . . , b_{2023}$ in the following way: $$b_1 = a_1 + a_2 + ... + a_{2023}$$$$b_2 =\sum_{1\le i_1 < i_2 \le2023} a_{i_1}a_{i_2} = a_1a_2 + a_1a_3 + a_1a_4 + ... + a_{2022}a_{2023}$$$$b_3 =\sum_{1\le i_1<i_2<i_3 \le2023} a_{i_1}a_{i_2}a_{i_3} = a_1a_2a_3 + a_1a_2a_4 + a_1a_2a_5 + ... + a_{2021}a_{2022}a_{2023}$$$$...$$$$b_{2023 }= a_1a_2a_3...a_{2023}$$It is known that $b_1, b_2, . . . , b_{2023}$3 are all negative. Amongst $a_1, a2, . . . a_{2023}$, $n$ of them are positive. Enter $n$ mod $1000$.

$a$ and $b$ are positive integers such that $a^3b = 14 \cdot 15!$ and the minimum possible value of $3a + b$ is $k$. Enter $k$ mod $1000$ .

An archer shoots $2023$ arrows with each having $\frac12$ probability of hitting the target. The probability of the total number of hits being divisible by $4$ can be expressed in the form of $\frac14 + x^{-y}$, where $x$ and $y$ are positive integers and $x$ is not a positive power of an integer other than itself. Enter $y-10x$.

For $f(x) \in Z[x]$, polynomials with integer coefficients, we define the “symbolic derivative”, $f'(x)$ of $f(x)$. For constant functions $f(x) = c$ we define $f'(x) = 0$ and if $f(x) = a_0 + a_1x + a_2x^2 + ...+ a_nx^n$ then we define $f'(x) = a_1 +2a_2x+...+na_nx^{n-1}$. Further, for $f(x) = a_0 +a_1x+...+a_nx^n$ we define a function $h : Z[x] \to N$ by $h(f(x)) = \sum^{n}_{i=0}|a_i|$, which we call the height of $f(x)$. If the number of polynomials $f(x)$ with integer coefficients satisfying $f'(f(x))f'(x) = 4xf(x)$ and $h(f(x)) < 23$ is $n$, enter $n$.

Let $f : R \to R$ be a continuous function such that for all real numbers $x$ and $y$, $$f(x+y)-f(x)-f(y) = xy+x^2y+xy^2$$and $f(6) = 69$. If $f(2\sqrt6) = a+b\sqrt{c}$, where $a$ and $b$ are integers and $c$ is a square-free integer, enter $a+b+c$.

A frog can “jump around” a point in the following way: if a frog at point $M$ jumps around point $N$, it lands on the point of rotation of $M$ that is $120^o$ anti-clockwise about $N$ (i.e. if the point of landing is $M'$, then $NM = NM'$ and $\angle MNM' = 120^o$ in directed angle). A frog is initially at point $X$. $A$, $B$, $C$ are points such that $X$, $A$, $B$ and $C$ are on the same plane and $XA = 10$, $AB = 15$. The frog first jumps around $A$, then around $B$, then around $C$, and then continues to jump over around $A$, $B$ and $C$ alternately. After $420$ jumps, the frog return the its original position at point $X$. The maximum possible value of $XC$ is $p$ and the minimum possible value of $XC$ is $q$. Enter $p^2 - q^2$.

Consider words made with letters $A_1, . . . ,A_{2023}$. We call a word humble if and only if it contains an even number of $A_1$’s and an odd number of $A_2$’s. Let $s$ be the number of humble words of length $2023$. Enter $s$ mod $1000$.

$\vartriangle ABC$ is a triangle such that $BC = 13$, $CA = 14$, $AB = 15$. $D, E, F$ are points on line segments $BC$, $CA$, $AB$ respectively such that $\vartriangle AEF$, $\vartriangle BFD$, $\vartriangle CDE$ have the same inradii. Let the inradii of these three triangles be $r_1$ and the inradius of $\vartriangle DEF$ be $r_2$. It is known that $r^2_1 + r^2_2 =\frac{ 25}{2}$ and $r_1 = \frac{p}{q} \le r_2$, where $p$, $q$ are integers that are coprime. Enter $p^2 + q^2$.

Let $\vartriangle ABC$ be an equilateral triangle with side length $12$. $D, E, F$ are points such that $ADEF$ is a square. The area of the possible geometric location of the centre of the square $ ADEF$ such the both segments $ED$ and $EF$ intersect the segment $BC$ (the possible geometric location of the centre is not a line or a curve) is $a\pi + b - c\sqrt{d}$, where $a$, $b$, $c$ and $d$ are positive integers and $d$ is square-free. Enter $a+b+c+d$.

Let $s$ be the number of positive numbers that satisfy $$x\lfloor x\rfloor \{x\} + x = 4^{2023}.$$Enter $s$ mod $1000$. Note: $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$, and $\{x\}$ denotes the fractional part of $x$, i.e. $x - \lfloor x\rfloor$.

King Cnut of the North Sea Empire wishes to see his Jomsvikings, a special viking force, train on a field next to his palace in Denmark. His military advisor, Godwin of Essex, wants to keep most of the Jomsvikings stationed in England where Godwin currently resides. Cnut has a strict requirement, $23$ of the vikings must all be training at most within $6$ feet of each other (i.e. the pairwise distance between them is at most $6$ feet) and if not, then $23$ of the vikings must all be training more than $3$ feet apart from each other. Assume the vikings are points on a plane; they have no areas. Godwin cannot get the vikings to coordinate and achieve this, but wants to send as few as possible. Let the minimum number of vikings that need to be sent to guarantee Cnut’s requirement is satisfied be $n$, enter $n$.

Pepe and Lintz the pig play a game. (a) In the first round, Lintz writes a degree 2023 polynomial on a blackboard where all the coefficients of the polynomial (including the constant term) are non-zero real numbers. Pepe then chooses two terms of the polynomial and exchanges their coefficients. If the resulting polynomial has no real rational roots, then Pepe wins. Otherwise, Lintz wins. Who has a winning strategy for this round and what is it? (b) The second round has the same rules as the first round, except that in this round, if the resulting polynomial has one or more negative real roots, then Pepe wins. Otherwise, Lintz wins. Who has a winning strategy for this round and what is it?

Let $b = b_kb_{k-1}...b_2b_1$ be a binary string (string that consists of $1$’s and $0$’s only) of length $k$, where $k$ is a positive integer. Based on this binary string we generate a two-dimensional “path” $a_0 \to a1 \to ... \to a_k$ such that $$a_0 = (0, 0),\,\,\, a_1 =\begin{cases}(1, 0), \,\,\, if \,\,\,b_1 = 1 \\ (-1, 0)\,\,\, if \,\,\,b_1 = 0 \end{cases}$$$$_{a_{n+1}}=\begin{cases} a_n + (1, 0),\,\,\, if \,\,\, b_{n+1} = 1, b_n = 0 \\ a_n + (0, 1),\,\,\, if \,\,\,b_{n+1} = 1, b_n = 1 \\ a_n - (1, 0),\,\,\, if \,\,\,b_{n+1} = 0, b_n = 1 \\ a_n - (0, 1),\,\,\, if \,\,\,b_{n+1} = 0, b_n = 0\end{cases} \,\,\, \,\,\, for \,\,\, n \in \{2, ..., k-1\}$$ We call a binary string $b$ ideal if and only if the path generated by $b$ ends in $(0, 0)$. For example $b = 1100$ is ideal, because it generates the following path: $$\underbrace{(0, 0)}_{a_0} \to \underbrace{(-1, 0)}_{a_1} \to \underbrace{(-1,-1) }_{a_2} \to \underbrace{(0,-1) }_{a_3} \to \underbrace{(0, 0) }_{a_4} .$$Let $S(k)$ be the set of all ideal binary strings of length $k$. For each positive integer $k$, characterize all elements of $S(k)$ and find an explicit formula for the number of elements in the set $S(k)$.

Let $f(n)$ be the product of non-zero digits of positive integer $n$ in base $10$ and let $f(0) = 1$. For example, $f(420) = 4 \cdot 2 = 8$. Also, let $s(n) =\sum^n_{i=0}f(i)$. Find the maximum value of positive integer $k$ such that for every non-negative integer $n$, $10^{2023} | n + 1$ implies $k | s(n)$. Note: For positive integers $a$ and $b$, $a | b$ means that $a$ is a divisor of $b$.

Let $\vartriangle ABC$ be a scalene (i.e. non-isosceles) triangle with circumcentre $O$. Let $J_A$, $J_B$, $J_C$ be the $A$-, $B$-, $C$-excentre of $\vartriangle ABC$ respectively. $OJ_A$ intersects $BC$ at $X$, $OJ_B$ intersects $CA$ at $Y$ , $OJ_C$ intersects $AB$ at $Z$. Prove that $AX$, $BY$ , $CZ$ and the Euler line of $\vartriangle ABC$ are concurrent. Note: The $A$-excentre of $\vartriangle ABC$ is the centre of the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond C. The $B$-, $C$-excentre of $\vartriangle ABC$ are defined similarly. The Euler line of a triangle is the line which passes through the orthocentre and the circumcentre of the triangle.

Find all functions $f : R \to R$ such that $$f(xf(y) + f(f(y))) = (x + 1)f(y)$$for all real numbers $x$ and $y$.

Let $\vartriangle ABC$ be an acute-angled scalene (i.e. non-isosceles) triangle. $D$ and $E$ are the foot of $B$-altitude and C-altitude of $\vartriangle ABC$ respectively. $M$ is the midpoint of $BC$. The circumcircles of $\vartriangle BEM$ and $\vartriangle CDM$ meet at $F$ (other than $M$). $X$ and $Y$ are the points of projection of $F$ on $BD$ and $CE$ respectively. Prove that $AF$ and $XY$ meet at the circumcircle of $\vartriangle DEM$.

Pepe has a large garden for frogs. He plays a game with Lintz the pig in the garden. The rules are as follows: 1. A positive real number $s$ is randomly generated by a computer system, called FROG (short form for “Fast, Random, Outstanding Generator of numbers”). 2. Lintz then draws a circle $K_1$ in Pepe’s garden (the garden is assumed to be an Euclidean plane). 3. Pepe then draws another circle $K_2$ in his garden. 4. After that, Lintz colours a finite number of arcs of $K_1$ in yellow, such that the total length of yellow arcs is more than two-third of the circumference of $K_1$. He also colours a finite number of arcs of $K_2$ in purple, such that the total length of purple arcs is more than two-third of the circumference of $K_2$. 5. Pepe puts three frogs, $A_1$, $B_1$, $C_1$ on the yellow arcs of $K_1$. He also puts three frogs, $A_2$, $B_2$, $C_2$ on the purple arcs of $K_2$. (Assume that the frogs are points, i.e. they have no areas.) Define $m_{A_1}$ , $m_{B_1}$ , $m_{C_1}$ , $h_{A_1}$ , $h_{B_1}$ , $h_{C_1}$ as the lengths of the $A$-median, $B$-median, $C$-median, $A$-altitude, $B$-altitude and $C$-altitude of $\vartriangle A_1B_1C_1$ respectively. $m_{A_2}$ , $m_{B_2}$ , $m_{C_2}$ , $h_{A_2}$ , $h_{B_2}$ , $h_{C_2}$ are defined similarly for $\vartriangle A_2B_2C_2$. Pepe wins if $$m^2_{A_2} (m^2_{B_1} + m^2_{C_1}- m^2_{A_1}) + m^2_{B_2} (m^2_{C_1} + m^2_{A_1}- m^2_{B_1}) + m^2_{C_2} (m^2_{A_1} + m^2_{B_1}- m^2_{C_1})$$$$+ \frac{1}{h^2_{A_2}}\left( \frac{1}{h^2_{B_1}}+\frac{1}{h^2_{C_1}}-\frac{1}{h^2_{A_1}}\right) + \frac{1}{h^2_{B_2}}\left( \frac{1}{h^2_{C_1}}+\frac{1}{h^2_{A_1}}-\frac{1}{h^2_{B_1}}\right) +\frac{1}{h^2_{C_2}}\left( \frac{1}{h^2_{A_1}}+\frac{1}{h^2_{B_1}}-\frac{1}{h^2_{C_1}}\right)= s$$Otherwise, Lintz wins. Find the range of $s$ where Pepe has a winning strategy and the range of $s$ where Lintz has a winning strategy. Prove your claim. PSI added last equation as an attachment also because it had too many indices, in case I mistyped any of them.