Indonesia Regional [hide=MO]also know as provincial level, is a qualifying round for National Math Olympiad[/hide]  
Year 2007  [hide=Part A]Part B consists of 5 essay / proof problems, posted [url=https://artofproblemsolving.com/community/c4h2684679p23289059]here[/url][/hide]

Time: 90 minutes [hide=Rules]
$\bullet$ Write only the answers to the questions given. 
$\bullet$ Some questions can have more than one correct answer. You are asked to provide the most correct or exact answer to a question like this. Scores will only be given to the giver of the most correct or most exact answer.
$\bullet$ Each question is worth 1 (one) point.[/hide]


p1. The largest $4$-digit odd number who is sum of all prime numbers is ...

p2. The amount of money consists of coins in the $500$ Rp.,  $200$ Rp. , and $100$ Rp. with a total value of  $100,000$ Rp.. If the $500$'s note is half the $200$'s note, but three times the $100$'s note, then the number of coins is ...

p3. The length of the hypotenuse of a right triangle is equal to twice the length of the shortest side, while the length of the third side is $ 1$ unit longer than the length of the shortest side. The area of ​​the triangle is ... square units .

p4. Among the numbers $2006$, $2007$ and $2008$, the number that has the most different prime factors is ...

p5. A used car dealer sells two cars for the same price. He loses $10\%$ on the first car, but breaks even (return on capital) for both cars. The merchant's profit percentage for the second car is

p6. Dona arranges five congruent squares into a flat shape. No square overlaps another square. If the area of ​​the figure obtained by Dona is $245$ cm$^2$, the perimeter of the figure is at least ... cm.

p7. Four football teams enter a tournament. Each team plays against each of the other teams once. Each time a team competes, a team gets $3$ points if it wins, $0$ if it loses and $ 1$ if the match ends in a draw. At the end of the tournament one of the teams gets a total score of $4$. The sum of the total points of the other three teams is at least ...

p8. For natural numbers $n$. In its simplest form, $1!1+2!2+3!3+...+n!n=...$

p9. Point $P$ is located in quadrant I on the line $y = x$. The point $Q$ lies on the line $y = 2x$ such that $PQ$ is perpendicular to the line $y = x$ and $PQ = 2$. Then the coordinates of $Q$ are ...

p10. The set of all natural numbers $n$ such that $6n + 30$ is a multiple of $2n + 1$ is ...

p11. The term constant in the $\left(2x^2-\frac{1}{x}\right)^9$ expansion is ...

p12. The abscissa of the point where the line $\ell$ intersects the $x$-axis and the coordinates of the point where it intersects $\ell$ with the $y$-axis are prime numbers. If $\ell$ also passes through the point $(3, 4)$, the equation of $\ell$ is ...

p13. Seventeen candies are packed into bags so that the number of candies in each two bags is at most $ 1$. The number of ways to pack the candies into at least two bags is ...

p14. If the maximum value of $x + y$ in the set $\{(x, y)|x\ge 0, y\ge0, x + 3y \le 6, 3x + y \le a\}$ is $4$, then $a = ...$

p15. A $5  \times 5 \times 5$ cube is composed of $125$ unit cubes. The surface of the large cube is then painted. The ratio of the sides (surfaces) of the $125$ painted to unpainted unit cubes is ....

p16. A square board is divided into $4 \times 4$ squares and colored like a chessboard. Each tile is numbered from $ 1$ to $16$. Andi wants to cover the tiles on the board with $7$ cards of $2 \times 1$ tile. In order for his $7$ cards to cover the board, he must discard two squares. The number of ways he disposes of two squares is ...

p17. The natural numbers $1, 2, ... , n$ are written on the blackboard, then one of the numbers is deleted. The arithmetic mean of the numbers left behind is $35 \frac{7}{17}$. The $n$-th possible number for this to happen is ...

p18. Given a triangle $ABC$ right at $A$, point $D$ at $AC$ and point $F$ at $BC$. If $AF\perp BC$ and $BD = DC = FC = 1$, then $AC = ...$

p19. Among all the natural number solutions $(x, y)$ of the equation $\frac{x+y}{2}+\sqrt{xy}=54$, the solution with the largest $x$ is $(x, y) = ...$

p20. Let $V$ be the set of points on the plane with integer coordinates and $X$ be the set of midpoints of all pairs of points in the set $V$.  To ensure that any member of $X$ also has integer coordinates, the number of members of $V$ must be at least ...
1. Let the sides of the right triangle be \( s \), \( s+1 \), and \( 2s \), where \( s \) is the shortest side.
2. By the Pythagorean Theorem, we have:
   \[
   s^2 + (s+1)^2 = (2s)^2
   \]
3. Expanding and simplifying the equation:
   \[
   s^2 + s^2 + 2s + 1 = 4s^2
   \]
   \[
   2s^2 + 2s + 1 = 4s^2
   \]
   \[
   2s^2 - 2s - 1 = 0
   \]
4. Solving the quadratic equation using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   s = \frac{2 \pm \sqrt{4 - 4 \cdot 2 \cdot (-1)}}{4}
   \]
   \[
   s = \frac{2 \pm \sqrt{4 + 8}}{4}
   \]
   \[
   s = \frac{2 \pm \sqrt{12}}{4}
   \]
   \[
   s = \frac{2 \pm 2\sqrt{3}}{4}
   \]
   \[
   s = \frac{1 \pm \sqrt{3}}{2}
   \]
5. Since \( s \) is positive, we take the positive root:
   \[
   s = \frac{1 + \sqrt{3}}{2}
   \]
6. The area of the triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base is \( s \) and the height is \( s+1 \):
   \[
   \text{Area} = \frac{1}{2} \times s \times (s+1)
   \]
7. Substituting \( s = \frac{1 + \sqrt{3}}{2} \):
   \[
   \text{Area} = \frac{1}{2} \times \frac{1 + \sqrt{3}}{2} \times \left( \frac{1 + \sqrt{3}}{2} + 1 \right)
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{1 + \sqrt{3}}{2} \times \left( \frac{1 + \sqrt{3} + 2}{2} \right)
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{1 + \sqrt{3}}{2} \times \frac{3 + \sqrt{3}}{2}
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{(1 + \sqrt{3})(3 + \sqrt{3})}{4}
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{3 + 3\sqrt{3} + \sqrt{3} + 3}{4}
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{6 + 4\sqrt{3}}{4}
   \]
   \[
   \text{Area} = \frac{1}{2} \times \frac{6 + 4\sqrt{3}}{4}
   \]
   \[
   \text{Area} = \frac{6 + 4\sqrt{3}}{8}
   \]
   \[
   \text{Area} = \frac{3 + 2\sqrt{3}}{4}
   \]
8. Therefore, the area of the triangle is:
   \[
   \boxed{\frac{3}{4} + \frac{\sqrt{3}}{2}}
   \]