The problems are really difficult to find online, so here are the problems.

P1. It is known that two positive integers $m$ and $n$ satisfy $10n - 9m = 7$ dan $m \leq 2018$. The number $k = 20 - \frac{18m}{n}$ is a fraction in its simplest form.
a) Determine the smallest possible value of $k$.
b) If the denominator of the smallest value of $k$ is (equal to some number) $N$, determine all positive factors of $N$.
c) On taking one factor out of all the mentioned positive factors of $N$ above (specifically in problem b), determine the probability of taking a factor who is a multiple of 4.

I added this because my translation is a bit weird.
[hide=Indonesian Version] Diketahui dua bilangan bulat positif $m$ dan $n$ dengan $10n - 9m = 7$ dan $m \leq 2018$. Bilangan $k = 20 - \frac{18m}{n}$ merupakan suatu pecahan sederhana.
a) Tentukan bilangan $k$ terkecil yang mungkin.
b) Jika penyebut bilangan $k$ terkecil tersebut adalah $N$, tentukan semua faktor positif dari $N$.
c) Pada pengambilan satu faktor dari faktor-faktor positif $N$ di atas, tentukan peluang terambilnya satu faktor kelipatan 4.[/hide]

P2. Let the functions $f, g : \mathbb{R} \to \mathbb{R}$ be given in the following graphs.
[hide=Graph Construction Notes]I do not know asymptote, can you please help me draw the graphs? Here are its complete description:
For both graphs, draw only the X and Y-axes, do not draw grids. Denote each axis with $X$ or $Y$ depending on which line you are referring to, and on their intercepts, draw a small node (a circle) then mark their $X$ or $Y$ coordinates only (since their other coordinates are definitely 0).
Graph (1) is the function $f$, who is a quadratic function with -2 and 4 as its $X$-intercepts and 4 as its $Y$-intercept. You also put $f$ right besides the curve you have, preferably just on the right-up direction of said curve.
Graph (2) is the function $g$, which is piecewise. For $x \geq 0$, $g(x) = \frac{1}{2}x - 2$, whereas for $x < 0$, $g(x) = - x - 2$. You also put $g$ right besides the curve you have, on the lower right of the line, on approximately $x = 2$.[/hide]
Define the function $g \circ f$ with $(g \circ f)(x) = g(f(x))$ for all $x \in D_f$ where $D_f$ is the domain of $f$.
a) Draw the graph of the function $g \circ f$.
b) Determine all values of $x$ so that $-\frac{1}{2} \leq (g \circ f)(x) \leq 6$.

P3. The quadrilateral $ABCD$ has side lengths $AB = BC = 4\sqrt{3}$ cm and $CD = DA = 4$ cm. All four of its vertices lie on a circle. Calculate the area of quadrilateral $ABCD$.

P4. There exists positive integers $x$ and $y$, with $x < 100$ and $y > 9$. It is known that $y = \frac{p}{777} x$, where $p$ is a 3-digit number whose number in its tens place is 5. Determine the number/quantity of all possible values of $y$.

P5. The 8-digit number $\overline{abcdefgh}$ (the original problem does not have an overline, which I fixed) is arranged from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$. Such number satisfies $a + c + e + g \geq b + d + f + h$. Determine the quantity of different possible (such) numbers.


To solve this problem, we need to find all possible values of \( y \) given the constraints \( x < 100 \), \( y > 9 \), and \( y = \frac{p}{777} x \), where \( p \) is a 3-digit number with the tens digit being 5.

1. **Factorize 777**:
   \[
   777 = 3 \times 7 \times 37
   \]
   Therefore, \( x \) must be a factor of 777. The factors of 777 are \( 1, 3, 7, 21, 37, 111, 259, 777 \). Since \( x < 100 \), the possible values for \( x \) are \( 1, 3, 7, 21, 37 \).

2. **Determine \( p \) for each \( x \)**:
   - For \( x = 37 \):
     \[
     y = \frac{p}{21} \implies 21 \mid p
     \]
     Since \( p = \overline{A5B} \) and \( 21 \mid p \), \( p \) must be divisible by both 3 and 7.
     - Divisibility by 3: \( A + 5 + B \) must be divisible by 3.
     - Divisibility by 7: \( 2A + B + 1 \) must be divisible by 7.
     
     Solving these conditions:
     \[
     A + 5 + B \in \{1, 4, 7, 10, 13, 16\}
     \]
     \[
     2A + B + 1 \in \{7, 14, 21\}
     \]
     The valid pairs \((A, B)\) are \((2, 2), (3, 7), (6, 1), (7, 6)\), giving \( p \in \{252, 357, 651, 756\} \).
     \[
     y \in \left\{ \frac{252}{21}, \frac{357}{21}, \frac{651}{21}, \frac{756}{21} \right\} = \{12, 17, 31, 36\}
     \]

   - For \( x = 21 \):
     \[
     y = \frac{p}{37} \implies 37 \mid p
     \]
     Since \( p = \overline{A5B} \) and \( 37 \mid p \), \( p \) must be divisible by 37.
     The valid values of \( p \) are \( 259, 555, 851 \).
     \[
     y \in \left\{ \frac{259}{37}, \frac{555}{37}, \frac{851}{37} \right\} = \{7, 15, 23\}
     \]

   - For \( x = 7 \):
     \[
     y = \frac{p}{111} \implies 111 \mid p
     \]
     Since \( p = \overline{A5B} \) and \( 111 \mid p \), the valid value of \( p \) is \( 555 \).
     \[
     y = \frac{555}{111} = 5
     \]

   - For \( x = 3 \):
     \[
     y = \frac{p}{259} \implies 259 \mid p
     \]
     Since \( p = \overline{A5B} \) and \( 259 \mid p \), the valid value of \( p \) is \( 259 \).
     \[
     y = \frac{259}{259} = 1
     \]

3. **Filter valid \( y \) values**:
   We need \( y > 9 \), so we discard \( y = 5 \) and \( y = 1 \).

   The valid \( y \) values are:
   \[
   \{12, 15, 17, 23, 31, 36\}
   \]

The final answer is \(\boxed{6}\)