[b]p1.[/b] Find the smallest positive multiple of $9$ whose digits are all even.


[b]p2.[/b] Anika writes down a positive real number $x$ in decimal form. When Nat erases everything to the left of the decimal point, the remaining value is one-fifth of x. Find the sum of all possible values of $x$.


[b]p3.[/b] A star-like shape is formed by joining up the midpoints and vertices of a unit square, as shown in the diagram below. Compute the area of this shape.
[img]https://cdn.artofproblemsolving.com/attachments/4/8/923b1bf26f6e9872b596e8c81ad1872137f362.png[/img]

[b]p4.[/b] Benny and Daria are running a $200$ meter foot race, each at a different constant speed. When Daria finishes the race, she is $14$ meters ahead of Benny. The next time they race, Daria starts 14 meters behind Benny, who starts at the starting line. Both runners run at the same constant speed as in the first race. When Daria reaches the finish line, compute, in centimeters, how far she is ahead of Benny. 


[b]p5.[/b] In one semester, Ronald takes ten biology quizzes, earning a distinct integer score from $91$ to $100$ on each quiz. He notices that after the first three quizzes, the average of his three most recent scores always increased. Compute the number of ways Ronald could have earned the ten quiz scores.


[b]p6.[/b] Ant and Ben are playing a game with stones. They begin with $Z$ stones on the ground. Ant and Ben take turns removing a prime number of stones. Ant moves first. The player who is first unable to make a valid move loses. Find the sum of all positive integers $Z \le 30$ such that Ben can guarantee a win with perfect play.


[b]p7.[/b] Let $P$ be a point in a regular octagon as shown in the diagram below. The areas of three triangles are shown. Find the area of the octagon.
[img]https://cdn.artofproblemsolving.com/attachments/0/9/6fde77eeafd04614046292175e4b1411158e85.png[/img]

[b]p8.[/b] Find the number of ordered triples $(a, b, c)$ of nonnegative integers with $a \le b \le c$ for which $5a + 4b + 6c = 1200$.


[b]p9.[/b] Define $$f(x) = \begin{cases} 
2x \,\,\,\, ,\,\,\,\,  0 \le  x < \frac12 \\
2 - 2x \,\,\,\, , \,\,\,\,  \frac12 \le x \le 1 \end{cases}$$
Michael picks a real number $0 \le x  \le 1$. Michael applies $f$ repeatedly to $x$ until he reaches $x$ again. Find the number of real numbers $x$ for which Michael applies $f$ exactly $12$ times.


[b]p10.[/b] In $\vartriangle ABC$, let point $H$ be the intersection of its altitudes and let $M$ be the midpoint of side $BC$. Given that $BC = 4$, $MA = 3$, and $\angle HMA = 60^o$, find the circumradius of $\vartriangle ABC$.


PS. You should use hide for answers.  Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. **Problem 1:**
   To find the smallest positive multiple of $9$ whose digits are all even, we need to use the divisibility rule for $9$, which states that the sum of the digits of the number must be a multiple of $9$. Since all digits must be even, the smallest even digits are $0, 2, 4, 6, 8$. The smallest sum of even digits that is a multiple of $9$ is $18$ (since $2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18$). The smallest number with a sum of digits equal to $18$ and all digits even is $288$.

   Therefore, the smallest positive multiple of $9$ whose digits are all even is:
   \[
   \boxed{288}
   \]

2. **Problem 2:**
   Given that when Nat erases everything to the left of the decimal point, the remaining value is one-fifth of $x$, we can write:
   \[
   \{x\} = \frac{1}{5}x
   \]
   where $\{x\}$ denotes the fractional part of $x$. This implies:
   \[
   \{x\} = \frac{1}{5}(\lfloor x \rfloor + \{x\})
   \]
   Rearranging, we get:
   \[
   \{x\} = \frac{1}{5}\lfloor x \rfloor + \frac{1}{5}\{x\}
   \]
   \[
   \frac{4}{5}\{x\} = \frac{1}{5}\lfloor x \rfloor
   \]
   \[
   \lfloor x \rfloor = 4\{x\}
   \]
   Since $\lfloor x \rfloor$ is an integer, $\{x\}$ must be one of $0, 0.25, 0.5, 0.75$. These correspond to $x = 0, 1.25, 2.5, 3.75$ respectively. Summing the positive values:
   \[
   1.25 + 2.5 + 3.75 = 7.5
   \]

   Therefore, the sum of all possible values of $x$ is:
   \[
   \boxed{7.5}
   \]

3. **Problem 3:**
   To find the area of the star-like shape formed by joining the midpoints and vertices of a unit square, we note that the area of the square is $1$. The star-like shape is the area of the square minus the sum of the areas of the eight white triangles.

   Using coordinate geometry, we place the square with vertices at $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. The vertices of one of the white triangles are $(0,0)$, $(0.5,0)$, and $(0.4,0.2)$. Using the Shoelace formula, the area of this triangle is:
   \[
   \text{Area} = \frac{1}{2} \left| 0(0) + 0.5(0.2) + 0.4(0) - (0(0.5) + 0(0.4) + 0.2(0)) \right| = \frac{1}{2} \left| 0.1 \right| = 0.05
   \]
   Since there are eight such triangles, the total area of the white triangles is:
   \[
   8 \times 0.05 = 0.4
   \]
   Therefore, the area of the star-like shape is:
   \[
   1 - 0.4 = 0.6
   \]

   Therefore, the area of the star-like shape is:
   \[
   \boxed{\frac{3}{5}}
   \]

4. **Problem 4:**
   Given that Daria finishes the race 14 meters ahead of Benny, we can determine the ratio of their speeds. Let $v_D$ be Daria's speed and $v_B$ be Benny's speed. Since Daria finishes 14 meters ahead, we have:
   \[
   \frac{v_B}{v_D} = \frac{186}{200} = \frac{93}{100}
   \]
   In the next race, Daria starts 14 meters behind Benny. When Daria runs 214 meters, Benny runs:
   \[
   \frac{93}{100} \times 214 = 199.02 \text{ meters}
   \]
   Therefore, the distance in meters between Daria and Benny when Daria finishes is:
   \[
   214 - 199.02 = 14.98 \text{ meters}
   \]
   Converting this to centimeters:
   \[
   14.98 \times 100 = 1498 \text{ centimeters}
   \]

   Therefore, when Daria reaches the finish line, she is ahead of Benny by:
   \[
   \boxed{98} \text{ centimeters}
   \]

5. **Problem 5:**
   Let the 10 scores Ronald gets be $a_1, a_2, \ldots, a_{10}$ and partition them into the sets $\{a_1, a_4, a_7, a_{10}\}$, $\{a_2, a_5, a_8\}$, $\{a_3, a_6, a_9\}$. The given condition is equivalent to:
   \[
   a_1 < a_4 < a_7 < a_{10}, \quad a_2 < a_5 < a_8, \quad a_3 < a_6 < a_9
   \]
   Therefore, for any partition of $\{91, 92, \ldots, 100\}$ into these three sets, there is exactly one way to order the scores within each set. The number of ways to partition and order the scores is:
   \[
   \binom{10}{4} \binom{6}{3} \binom{3}{3} = 210 \times 20 \times 1 = 4200
   \]

   Therefore, the number of ways Ronald could have earned the ten quiz scores is:
   \[
   \boxed{4200}
   \]

6. **Problem 6:**
   Let $S$ be the set of positive integers no larger than $30$ for which Ben can guarantee a win if $Z = n$. First, note that $0$ and $1$ are losing positions because no move can be made for whoever's turn it is. Hence, we have $1 \in S$. It then follows that all prime numbers or one more than prime numbers are winning positions because removing that prime makes it a losing position for the opponent.

   After removing these numbers, we are left to check $9, 10, 15, 16, 21, 22, 25, 26, 27, 28$. For $Z = 9$, if Ant takes $2, 3, 5, 7$, Ben can respond with $7, 5, 3, 2$ respectively, which turns it into a losing position for Ant, so $9 \in S$. Similarly, any move Ant makes when $Z = 10$, Ben can respond with a move to turn it into a losing position for Ant, so $10 \in S$ as well.

   From here, since $9$ and $10$ are losing positions for whoever's turn it is, it follows that any number of the form $p + 9$ or $p + 10$ where $p$ is a prime are winning positions.

   After removing these numbers, we are only left to check $25$. For $Z = 25$, we proceed with casework on how many stones Ant removes on his first move:
   - **2:** Ben can take $23$ stones, leaving $0$ stones which is a losing position for Ant.
   - **3:** Ben can take $13$ stones, leaving $9$ stones which is a losing position for Ant.
   - **5:** Ben can take $19$ stones, leaving $1$ stone which is a losing position for Ant.
   - **7:** Ben can take $17$ stones, leaving $1$ stone which is a losing position for Ant.
   - **11:** Ben can take $13$ stones, leaving $1$ stone which is a losing position for Ant.
   - **13:** Ben can take $11$ stones, leaving $1$ stone which is a losing position for Ant.
   - **17:** Ben can take $7$ stones, leaving $1$ stone which is a losing position for Ant.
   - **19:** Ben can take $5$ stones, leaving $1$ stone which is a losing position for Ant.
   - **23:** Ben can take $2$ stones, leaving $1$ stone which is a losing position for Ant.

   Hence, whichever move Ant makes when $Z = 25$, Ben can respond with a move that will put Ant into a losing position. It follows that $25$ is a losing position for Ant, so $25 \in S$. Putting everything together, we have:
   \[
   S = \{1, 9, 10, 25\}
   \]
   Therefore, the sum of all positive integers $Z \le 30$ such that Ben can guarantee a win with perfect play is:
   \[
   1 + 9 + 10 + 25 = \boxed{45}
   \]