The test this year was held on Monday, 22 August 2022 on 09.10-11.40 (GMT+7) for the essay section and was held on 12.05-13.30 (GMT+7) for the short answers section, which was to be done in an hour using the Moodle Learning Management System. Each problem in this section has a weight of 2 points, with 0 points for incorrect or unanswered problems, whereas in the essay section each problem has a weight of 7 points. As with last year, calculators, protractors and set squares are prohibited. (Of course abacus is also prohibited, but who uses abacus?)

Anyway here are the problems.

[b]Part 1: Speed Round (60 minutes)[/b]
The problem was presented in no particular order, although here I will order it in a rough order of increasing difficulty.

[b]Problem 1.[/b] The number of positive integer solutions $(m,n)$ to the equation
\[ m^n = 17^{324} \] is $\ldots.$

[b][url=https://artofproblemsolving.com/community/c6h2909996_7_digits_numbers]Problem 2.[/url][/b] Consider the increasing sequence of all 7-digit numbers consisting of all of the following digits: $1,2,3,4,5,6,7$. The 2024th term of the sequence is $\ldots.$

[b]Problem 3.[/b] Suppose $(a,b)$ is a positive integer solution of the equation
\[ \sqrt{a + \frac{15}{b}} = a \sqrt{ \frac{15}{b}}. \] The sum of all possible values of $b$ is $\ldots.$

[b]Problem 4.[/b] It is known that $ABCD$ is a trapezoid such that $AB$ is parallel to $CD$, with the length of $AB = 6$ and $CD = 7$. It is known that points $P$ and $Q$ are on $AD$ and $BC$ respectively such that $PQ$ is parallel to $AB$. If the perimeter of trapezoid $ABQP$ is the same as the perimeter of $PQCD$ and $AD + BC = 10$, then the length of $20PQ$ is $\ldots.$

[b]Problem 5.[/b] Suppose $ABC$ is a triangle with side lengths $AB = 16$, $AC=23$, and $\angle{BAC} = 30^{\circ}$. The maximum possible area of a rectangle whose one of its sides lies on the line $BC$, and the two other vertices each lie on $AB$ and $AC$ is $\ldots.$

[b]Problem 6.[/b] Suppose $a,b,c$ are natural numbers such that $a+2b+3c=73$. The minimum possible value of $a^2+b^2+c^2$ is $\ldots.$

[b]Problem 7.[/b] An equilateral triangle with a side length of $21$ is partitioned into $21^2$ unit equilateral triangles, and the sides of the small equilateral triangles are all parallel to the original large triangle. The number of paralellograms which are made up of the unit equilateral triangles is $21k$. Then the value of $k = \ldots.$

[b]Problem 8.[/b] Define the sequence $\{a_n\}$ with $a_1 > 3$, and for all $n \geq 1$, the following condition:
\[ 2a_{n+1} = a_n(-1 + \sqrt{4a_n - 3}) \] is satisfied. If $\vert a_1 - a_{2022} \vert = 2023$, then the value of
\[ \sum_{i=1}^{2021} \frac{a_{i+1}^3}{a_i^2 + a_i a_{i+1} + a_{i+1}^2} = \ldots. \]

[b]Problem 9.[/b] Suppose $P(x)$ is an integer polynomial such that $P(6)P(38)P(57) + 19$ is divisible by $114$. If $P(-13) = 479$, and $P(0) \geq 0$, then the minimum value of $P(0)$ is $\ldots.$

[b]Problem 10.[/b] The number of [b]nonempty[/b] subsets of $S = \{1,2, \ldots, 21\}$ such that the sum of its elements is divisble by 4 is $2^k - m$; where $k, m \in \mathbb{Z}$ and $0 \leq m < 2022$. The value of $10k + m$ is $\ldots.$

To solve the problem, we need to minimize the expression \(a^2 + b^2 + c^2\) given the constraint \(a + 2b + 3c = 73\) where \(a, b, c\) are natural numbers. We will use the Cauchy-Schwarz inequality to find a lower bound for \(a^2 + b^2 + c^2\).

1. **Applying the Cauchy-Schwarz Inequality:**
   \[
   (1^2 + 2^2 + 3^2)(a^2 + b^2 + c^2) \geq (a \cdot 1 + b \cdot 2 + c \cdot 3)^2
   \]
   Simplifying the left-hand side:
   \[
   (1 + 4 + 9)(a^2 + b^2 + c^2) = 14(a^2 + b^2 + c^2)
   \]
   Simplifying the right-hand side:
   \[
   (a + 2b + 3c)^2 = 73^2 = 5329
   \]
   Therefore, we have:
   \[
   14(a^2 + b^2 + c^2) \geq 5329
   \]
   Dividing both sides by 14:
   \[
   a^2 + b^2 + c^2 \geq \frac{5329}{14} \approx 380.642857
   \]
   Since \(a, b, c\) are natural numbers, \(a^2 + b^2 + c^2\) must be an integer. Thus:
   \[
   a^2 + b^2 + c^2 \geq 381
   \]

2. **Checking if the Equality Holds:**
   The equality in the Cauchy-Schwarz inequality holds if and only if:
   \[
   \frac{a}{1} = \frac{b}{2} = \frac{c}{3} = k
   \]
   for some real number \(k\). This implies:
   \[
   a = k, \quad b = 2k, \quad c = 3k
   \]
   Substituting these into the constraint \(a + 2b + 3c = 73\):
   \[
   k + 2(2k) + 3(3k) = 73
   \]
   \[
   k + 4k + 9k = 73
   \]
   \[
   14k = 73
   \]
   \[
   k = \frac{73}{14} \approx 5.214
   \]
   Since \(k\) is not an integer, we cannot use this exact value. We need to find natural numbers \(a, b, c\) that satisfy the constraint and minimize \(a^2 + b^2 + c^2\).

3. **Finding Natural Numbers \(a, b, c\):**
   We test values close to \(k = 5.214\):
   - Let \(a = 5\), \(b = 10\), and \(c = 16\):
     \[
     a + 2b + 3c = 5 + 2(10) + 3(16) = 5 + 20 + 48 = 73
     \]
     \[
     a^2 + b^2 + c^2 = 5^2 + 10^2 + 16^2 = 25 + 100 + 256 = 381
     \]

Thus, the minimum value of \(a^2 + b^2 + c^2\) is 381.

The final answer is \(\boxed{381}\)