p1. Bahri lives quite close to the clock gadang in the city of Bukit Tinggi West Sumatra. Bahri has an antique clock. On Monday $4$th March $2013$ at $10.00$ am, Bahri antique clock is two minutes late in comparison with Clock Tower. A day later, the antique clock was four minutes late compared to the Clock Tower. March $6$, $2013$ the clock is late six minutes compared to Jam Gadang. The following days Bahri observed that his antique clock exhibited the same pattern of delay. On what day and what date in $2014$ the antique Bahri clock (hand short and long hands) point to the same number as the Clock Tower?


p2. In one season, the Indonesian Football League is participated by $20$ teams football. Each team competes with every other team twice. The result of each match is $3$ if you win, $ 1$ if you draw, and $0$ if you lose. Every week there are $10$ matches involving all teams. The winner of the competition is the team that gets the highest total score. At the end what week is the fastest possible, the winner of the competition on is the season certain?


p3. Look at the following picture. The quadrilateral $ABCD$ is a cyclic. Given that $CF$ is perpendicular to $AF$, $CE$ is perpendicular to $BD$, and $CG$ is perpendicular to $AB$. Is the following statements true?
Write down your reasons. $$\frac{BD}{CE}=\frac{AB}{CG}+ \frac{AD}{CF}$$
[img]https://cdn.artofproblemsolving.com/attachments/b/0/dbd97b4c72bc4ebd45ed6fa213610d62f29459.png[/img]


p4. Suppose $M=2014^{2014}$. If the sum of all the numbers (digits) that make up the number $M$ equals $A$ and the sum of all the digits that make up the number $A$ equals $B$, then find the sum of all the numbers that make up $B$.


p5. Find all positive integers $n < 200$ so that $n^2 + (n + 1)^2$ is square of an integer.
To solve the problem, we need to find the sum of all the digits that make up \( B \), where \( B \) is the sum of the digits of \( A \), and \( A \) is the sum of the digits of \( M = 2014^{2014} \).

1. **Estimate the number of digits in \( M \):**
   \[
   M = 2014^{2014}
   \]
   We can approximate \( 2014 \) as slightly less than \( 2048 \), which is \( 2^{11} \). Therefore:
   \[
   2014 < 2048 = 2^{11}
   \]
   Thus:
   \[
   2014^{2014} < (2^{11})^{2014} = 2^{22154}
   \]
   Since \( 2^{10} \approx 10^3 \), we have:
   \[
   2^{22154} \approx 10^{22154 \cdot \log_{10}(2)} \approx 10^{22154 \cdot 0.301} \approx 10^{6678.254}
   \]
   Therefore, \( M \) has approximately 6679 digits.

2. **Sum of the digits of \( M \):**
   The maximum sum of the digits of a number with 6679 digits is:
   \[
   9 \times 6679 = 60111
   \]
   Thus, \( A \leq 60111 \).

3. **Sum of the digits of \( A \):**
   The maximum sum of the digits of \( A \) (which has at most 5 digits) is:
   \[
   9 \times 5 = 45
   \]
   Thus, \( B \leq 45 \).

4. **Sum of the digits of \( B \):**
   The maximum sum of the digits of \( B \) (which has at most 2 digits) is:
   \[
   9 + 9 = 18
   \]

5. **Modulo 9 properties:**
   We use the property that the sum of the digits of a number is congruent to the number modulo 9. Therefore:
   \[
   2014 \equiv 2 + 0 + 1 + 4 = 7 \pmod{9}
   \]
   \[
   2014^{2014} \equiv 7^{2014} \pmod{9}
   \]
   Since \( 7^3 \equiv 1 \pmod{9} \):
   \[
   7^{2014} = (7^3)^{671} \cdot 7 \equiv 1^{671} \cdot 7 \equiv 7 \pmod{9}
   \]
   Therefore:
   \[
   M \equiv 7 \pmod{9}
   \]
   \[
   A \equiv 7 \pmod{9}
   \]
   \[
   B \equiv 7 \pmod{9}
   \]

Since \( B \) is a single-digit number, the sum of its digits is simply \( B \). Therefore, the sum of all the digits that make up \( B \) is \( 7 \).

The final answer is \( \boxed{7} \).