$a_1, a_2, ..., a_8$ are $8$ distinct positive integers. $b_1, b_2, ..., b_8$ are another $8$ distinct positive integers $(a_i, b_j$ are not necessarily distinct for $i, j = 1, 2, ...8)$. Enter the smallest possible value of $a^2_1b_1 + a^22b_2 + ... + a^2_8b_8$.

A frog can jump over a point in the following way: if a frog at point $M$ jumps over point $N$, it lands on the point of reflection of point $M$ over $N$. A frog is initially at point $X$. $A$, $B$, $C$ are points such that $X$, $A$, $B$ and $C$ are on the same plane. The frog first jumps over $A$, then over $B$, then over $C$, and then continues to jump over point $A$, $B$ and $C$ alternately. Suppose $XA = 36$ cm, $XB = 84$ cm, $XC = 48$ cm. After $2021$ jumps, the frog is at point $Y$ . $XY = k$ cm. Enter $k$.

Enter the smallest positive integer $n$ such that any set of $n$ relatively prime integers strictly greater than $1$ and strictly less than $2022$ contains at least $1$ prime number.

We define $a(n)$ as the number of positive divisors of $n$ and $b(n)$ as the sum of the positive divisors of $n$. There are $m$ positive integers $n$ between $1$ and $2022$ inclusive such that both $a(n)$ and $b(n)$ are odd. Enter $m$.

We call a coloring of a $2022 \times 2022$ checkerboard a banger coloring if it is colored by $2$ colors, black and white, such that each $2 \times 2$ square contains an even number of black cells. The number of banger colorings is $b$. Let $b = b_1b_2... cd$ (so $d$ is the last digit of $b$ and $c$ is the $2$nd last digit of $b$). Enter $c^2 + d^2 + 5$.

Let $w$ be an arbitrary complex number with a non-zero imaginary part and a non-zero real part. There are $n$ values of $k \in \{0, 1, 2, 3, 4, . . . , 999\}$ for which $$z =\left( \overline{w}w^3 - w \overline{w}^3\right)^k$$is a real number. Enter $n$.

Let $s(n)$ be defined as the sum of digits of $n$. Enter the sum of all positive integers $n$ for which $$s(11^n) = 2^n,$$

A subset of $\{1, 2, ..., 12\}$ is called trivoidant if no two elements of the subset differ by $3$. Enter the number of trivoidant subsets of $\{1, 2, ..., 12\}$.

Let $n$ be a positive integer and $s(n)$ be defined as the sum of digits of $n$. Also, let $$q(n) :=\frac{s(2n)}{s(n)}.$$If $a$ is the smallest possible value of $q(n)$ and $b$ is the biggest possible value of $q(n)$, then enter $100(a + b).$

There are n points in a $3$-dimensional space. If two points are exactly $1$ unit from each other, a line is joint between them. Let the number of total lines being joint be $\ell$. Suppose the maximum possible value of $\frac{\ell}{ n^2}$ is $\frac{p}{q}$ , where $p$ and $q$ are positive integers that are relatively prime to each other. Enter $100p + q$.

Let $f(n)$ denote the largest positive integer $m$ such that $7^m$ divides $H(n)$ where $H(n) = 1^1 \cdot 2^2\cdot \cdot \cdot n^n$. As $n$ gets arbitrarily large, $f(n) \sim kn^2$. Enter $a^2 + b^2$ where $k = \frac{a}{b}$ , $a, b$ are positive integers and gcd$(a, b) = 1$. For those unfamiliar of $\sim$, equivalently $$k = \lim_{n \to \infty} \frac{f(n)}{n^2}$$

Let $\omega$ be the incircle of $\triangle ABC$. Suppose $\omega$ touches $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. $AD$, $BE$, $CF$ intersect with $\omega$ at points $G$, $H$, $I$ respectively. $HI$ and $AD$ intersect at point $X$. If $GH = 8$, $HI = 9$, $IG = 10$, $IX =\frac{p}{q}$, where $p$, $q$ are positive integers that are relatively prime to each other. Enter $p + q$.

$30$ people stand on a circle with radius of $1$ unit. In view of the COVID-$19$ outbreak, the government has implemented social distancing policy. The distance between $2$ people should be more than $1$ unit, otherwise they will be fined. The amount of fine for each person is equal to $\$$(number of people that are within $1$ unit from him/her). For example, if $3$ people have distances of less than or equal to $1$ unit from person $A$, then person $A$ has to pay a fine of $\$3$. Enter the minimum total fine (in dollars) that has to be paid by all the $30$ people. (Assume that the peoples are points, ie they have no areas)

For real numbers $\theta_1, \theta_2, ...,\theta_11$ satisfying $$\sum^{11}_{i=1}\sin(\theta_i) = 1.$$The minimum value of $$\sum^{11}_{i=1}\sin(3\theta_i)$$is $-\frac{a}{b}$ where a, b are positive integers and gcd$(a, b) = 1$. Enter $a - b$.

Let $D$ be the centre of circle which passes through the mid-points of the three edges of $\triangle ABC$, $E$ and $F$ be points on $AB$ and $AC$ respectively such that $DE\perp AB$ and $DF\perp AC$. It is known that $AB = 15$, $AC = 20$, $4DE = 3DF$. $BC^2 = q+r\sqrt{s}$, where $q$ and $r$ are integers and $s$ is a square free integer. Enter $q + r + s$.

Find all pairs $(n,m)$ of positive integers such that $$9^n -5 \cdot 2^m = 1$$Show that you have found all such pairs.

For $k \ge 3$, $P(k, n)$ denotes the number of dots that create a regular $k$-gon with side length $n$. For example, here are the few first triangular numbers $P(3, n)$ and the first few square numbers $P(4, n)$ respectively: Here are the first few pentagonal numbers $P(5, n)$: (a) Find all arithmetic sequences of the form $$(P(k, n), P(k, n + 1), P(k + 1, n + 1))$$ (b) For $a \in Z$, show that there are infinitely many arithmetic sequences of the form $$(P(k, n), P(k, n + a), P(k + a, n + a))$$ Note: $n$ is a positive integer here, a member of the set $\{1, 2, 3,...\}$.

Define $A_1 = \emptyset$ and $B_1 = \{420\}$ to be our starting sets. We generate a sequence of sets as such $$A_{n+1} = \{x + 1 : x \in B_n \}$$$$B_{n+1} = (A_n \cup B_n) - (A_n \cap B_n)$$Show that $B_n = \{420\}$ if and only if $n$ is a power of $2$. Note: $\emptyset$ denotes the empty set, $\cup$ denotes set union, $\cap$ denotes set intersection and $-$ denotes set difference.

Let $ABC$ be a non-isosceles acute angle triangle. $A_0$ is the point of intersection of the $A$-symmedian and the circumcircle of $\vartriangle ABC$ (other than $A$). $A_1$ is the mid-point of $AA_0$. $A_2$ is the point of reflection of $A_0$ over the side opposite to $A$ (which is $BC$). $B_0$, $B_1$, $B_2$, $C_0$, $C_1$, $C_2$ are defined similarly. $O_1$, $O_2$, $N$ are the circumcentres of $\vartriangle A_1B_1C_1$, $\vartriangle A_2B_2C_2$ and the triangle formed by the midpoints of the three sides of $\vartriangle ABC$, respectively. $M$ is the midpoint of the centroid and the symmedian point of $\vartriangle ABC$. Prove that $O_1MO_2N$ is a parallelogram.

Let's take any positive integer $n_0 \in \{1, 2,...,\}$. Then, let $n_1$ be equal to the sum of digits of the square of $n_0$. Repeat this process to obtain $n_2$ from $n_1$, $n_3$ from $n_2$ and so on. In this way we obtain a sequence of natural numbers denoted as $(n_0; n_1, n_2,...)$. For example when we pick $n_0 = 7$ we obtain $$7 \to 4 + 9 = 13 \to 1 + 6 + 9 = 16 \to ...$$which creates sequence $$(7; 13, 16, ,,,)$$If in the $(n_0; n_1, n_2,...)$ sequence we are able to find two indices $i$ and $j$ such that $i > j$ and $n_i = n_j$ , then we say that this sequence "gets caught in a loop". If in the $(n_0; n_1, n_2,...)$ sequence we are able to find an index $i > 0$ such that $n_i = n_0$, then we call this sequence "a loop". Find all possible loops and prove that for every $n_0$, the sequence $(n_0; n_1, n_2,...)$ gets caught in a loop.

Let $O$ be a fixed point on a plane. $P_1$, $P_2$, $...$ , $P_{2022}$ are $2022$ variable points on the same plane which do not coincide with $O$. Initially, $P_1$ is an arbitrary point and the line segment $OP_{n+1}$ is $\frac{\pi}{1011}$ anticlockwise to the line segment $OP_n$ for $n = 1$, $2$,$...$, $2021$. In other words, $\angle P_nOP_{n+1} = \frac{\pi}{1011}$ in directed angles. A group of points is said to be "perfect" if $O$ is the point such that the sum of the distances from that point to all the points in the group is the smallest possible. For example, the group of points formed by $P_1$, $P_2$, $P_3$, $P_4$, $P_5$ is said to be "perfect" if $$OP_1 + OP_2 + OP_3 + OP_4 + OP_5 \le XP_1 + XP_2 + XP_3 + XP_4 + XP_5$$for any point $X$ on the plane. (a) Prove that the group of points formed by $P_1$, $P_2$, $...$, $P_{2022}$ is "perfect" initially. (b) You are given a task to remove the points $P_1$, $P_2$, $...$, $P_{2022}$ according the following rules. In each step, you could remove one point, and then rotate another point anticlockwise for $\frac{\pi}{3}$ around point $O$. After each step, the group formed by the remaining points on the plane must remain to be "perfect". Devise an algorithm to remove $1348$ points from the initial $2022$ points and prove that it works.

Let $n \in \{1, 2,...\}$ and $x, y, z$ all greater than $0$ be real numbers such that $x + y + z = 3$. Prove that $$(\sqrt{x} + 1)^n + (\sqrt{y} + 1)^n + (\sqrt{z} + 1)^n - n(xy + yz + zx) \ge 3(2^n - n)$$