p1. A fraction is called Toba-$n$ if the fraction has a numerator of $1$ and the denominator of $n$. If $A$ is the sum of all the fractions of Toba-$101$, Toba-$102$, Toba-$103$, to Toba-$200$, show that $\frac{7}{12} <A <\frac56$.


p2. If $a, b$, and $c$ satisfy the system of equations 
$$  \frac{ab}{a+b}=\frac12$$
$$\frac{bc}{b+c}=\frac13 $$
$$ \frac{ac}{a+c}=\frac17 $$
Determine the value of $(a- c)^b$.


p3. Given triangle $ABC$. If point $M$ is located at the midpoint of $AC$, point $N$ is located at the midpoint of $BC$, and the point $P$ is any point on $AB$. Determine the area of ​​the quadrilateral $PMCN$.
[img]https://cdn.artofproblemsolving.com/attachments/4/d/175e2d55f889b9dd2d8f89b8bae6c986d87911.png[/img]


p4. Given the rule of motion of a particle on a flat plane $xy$ as following:
$N: (m, n)\to (m + 1, n + 1)$
$T: (m, n)\to (m + 1, n - 1)$, where $m$ and $n$ are integers.
How many different tracks are there from $(0, 3)$ to $(7, 2)$ by using the above rules ?


p5. Andra and Dedi played “SUPER-AS”. The rules of this game as following. Players take turns picking marbles from a can containing $30$ marbles. For each take, the player can take the least a minimum of $ 1$ and a maximum of $6$ marbles. The player who picks up the the last marbels is declared the winner. If Andra starts the game by taking $3$ marbles first, determine how many marbles should be taken by Dedi and what is the next strategy to take so that Dedi can be the winner.
### Problem 1:
We need to show that the sum \( A \) of the fractions from Toba-101 to Toba-200 satisfies the inequality \( \frac{7}{12} < A < \frac{5}{6} \).

1. The sum \( A \) can be written as:
   \[
   A = \sum_{n=101}^{200} \frac{1}{n}
   \]

2. We can approximate this sum using the integral test for convergence of series. The harmonic series sum can be approximated by the natural logarithm:
   \[
   \sum_{n=101}^{200} \frac{1}{n} \approx \int_{101}^{201} \frac{1}{x} \, dx = \ln(201) - \ln(101)
   \]

3. Calculate the natural logarithms:
   \[
   \ln(201) \approx 5.3033 \quad \text{and} \quad \ln(101) \approx 4.6151
   \]
   \[
   \ln(201) - \ln(101) \approx 5.3033 - 4.6151 = 0.6882
   \]

4. Compare this with the given bounds:
   \[
   \frac{7}{12} \approx 0.5833 \quad \text{and} \quad \frac{5}{6} \approx 0.8333
   \]

5. Since \( 0.5833 < 0.6882 < 0.8333 \), we have:
   \[
   \frac{7}{12} < A < \frac{5}{6}
   \]

\(\blacksquare\)

### Problem 2:
We need to determine the value of \( (a - c)^b \) given the system of equations:
\[
\frac{ab}{a+b} = \frac{1}{2}, \quad \frac{bc}{b+c} = \frac{1}{3}, \quad \frac{ac}{a+c} = \frac{1}{7}
\]

1. From the first equation:
   \[
   \frac{ab}{a+b} = \frac{1}{2} \implies 2ab = a + b \implies 2ab - a - b = 0 \implies a(2b-1) = b
   \]
   \[
   a = \frac{b}{2b-1}
   \]

2. From the second equation:
   \[
   \frac{bc}{b+c} = \frac{1}{3} \implies 3bc = b + c \implies 3bc - b - c = 0 \implies b(3c-1) = c
   \]
   \[
   b = \frac{c}{3c-1}
   \]

3. From the third equation:
   \[
   \frac{ac}{a+c} = \frac{1}{7} \implies 7ac = a + c \implies 7ac - a - c = 0 \implies a(7c-1) = c
   \]
   \[
   a = \frac{c}{7c-1}
   \]

4. Equate the expressions for \( a \):
   \[
   \frac{b}{2b-1} = \frac{c}{7c-1}
   \]

5. Substitute \( b = \frac{c}{3c-1} \) into the equation:
   \[
   \frac{\frac{c}{3c-1}}{2\left(\frac{c}{3c-1}\right)-1} = \frac{c}{7c-1}
   \]
   \[
   \frac{c}{3c-1} \cdot \frac{3c-1}{2c-1} = \frac{c}{7c-1}
   \]
   \[
   \frac{c}{2c-1} = \frac{c}{7c-1}
   \]
   \[
   2c-1 = 7c-1 \implies 5c = 0 \implies c = 0
   \]

6. Since \( c = 0 \), substitute back to find \( a \) and \( b \):
   \[
   a = \frac{0}{7(0)-1} = 0, \quad b = \frac{0}{3(0)-1} = 0
   \]

7. Therefore, \( (a - c)^b = (0 - 0)^0 = 1 \).

\(\boxed{1}\)

### Problem 3:
We need to determine the area of the quadrilateral \( PMCN \).

1. Since \( M \) and \( N \) are midpoints of \( AC \) and \( BC \) respectively, \( M \) and \( N \) divide \( AC \) and \( BC \) into equal halves.

2. The area of triangle \( ABC \) is denoted as \( \Delta \).

3. The area of quadrilateral \( PMCN \) is half of the area of triangle \( ABC \) because \( M \) and \( N \) are midpoints, and \( P \) is any point on \( AB \).

4. Therefore, the area of \( PMCN \) is:
   \[
   \text{Area of } PMCN = \frac{1}{2} \Delta
   \]

\(\boxed{\frac{1}{2} \Delta}\)

### Problem 4:
We need to determine the number of different tracks from \( (0, 3) \) to \( (7, 2) \) using the given rules.

1. The rules \( N: (m, n) \to (m+1, n+1) \) and \( T: (m, n) \to (m+1, n-1) \) imply that each step increases \( m \) by 1 and either increases or decreases \( n \) by 1.

2. To go from \( (0, 3) \) to \( (7, 2) \), we need 7 steps in total (since \( m \) increases from 0 to 7).

3. Let \( x \) be the number of \( N \) steps and \( y \) be the number of \( T \) steps. We have:
   \[
   x + y = 7 \quad \text{and} \quad 3 + x - y = 2
   \]
   \[
   x - y = -1
   \]

4. Solving the system of equations:
   \[
   x + y = 7
   \]
   \[
   x - y = -1
   \]
   \[
   2x = 6 \implies x = 3
   \]
   \[
   y = 7 - 3 = 4
   \]

5. The number of different tracks is given by the binomial coefficient:
   \[
   \binom{7}{3} = \frac{7!}{3!4!} = 35
   \]

\(\boxed{35}\)

### Problem 5:
We need to determine how many marbles Dedi should take and the strategy to ensure Dedi wins.

1. Andra starts by taking 3 marbles, leaving 27 marbles.

2. To ensure Dedi wins, Dedi should always leave a multiple of 7 marbles after his turn.

3. If Dedi takes \( x \) marbles, then \( 27 - x \) should be a multiple of 7:
   \[
   27 - x \equiv 0 \pmod{7}
   \]
   \[
   27 \equiv x \pmod{7}
   \]
   \[
   27 \equiv 6 \pmod{7}
   \]
   \[
   x = 6
   \]

4. Dedi should take 6 marbles, leaving 21 marbles. After that, Dedi should always take \( 7 - y \) marbles if Andra takes \( y \) marbles, ensuring Dedi always leaves a multiple of 7 marbles.

\(\boxed{6}\)