[b]p19.[/b] $A_1A_2...A_{12}$ is a regular dodecagon with side length $1$ and center at point $O$. What is the area of the region covered by circles $(A_1A_2O)$, $(A_3A_4O)$, $(A_5A_6O)$, $(A_7A_8O)$, $(A_9A_{10}O)$, and $(A_{11}A_{12}O)$?
$(ABC)$ denotes the circle passing through points $A,B$, and $C$.


[b]p20.[/b] Let $N = 2000... 0x0 ... 00023$ be a $2023$-digit number where the $x$ is the $23$rd digit from the right. If$ N$ is divisible by $13$, compute $x$.


[b]p21.[/b] Alice and Bob each visit the dining hall to get a grilled cheese at a uniformly random time between $12$ PM and $1$ PM (their arrival times are independent) and, after arrival, will wait there for a uniformly random amount of time between $0$ and $30$ minutes. What is the probability that they will meet?


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let \( N' = 200\ldots0023 \), where there are \( 2020 \) zeroes. We need to find \( x \) such that \( N = N' + x \cdot 10^{2000} \) is divisible by \( 13 \).

2. First, we compute \( N' \mod 13 \):
   \[
   N' = 2 \cdot 10^{2022} + 23
   \]
   Since \( 10^3 \equiv -1 \pmod{13} \), we have:
   \[
   10^{2022} = (10^3)^{674} \equiv (-1)^{674} \equiv 1 \pmod{13}
   \]
   Therefore:
   \[
   2 \cdot 10^{2022} \equiv 2 \cdot 1 \equiv 2 \pmod{13}
   \]
   Adding the constant term:
   \[
   N' \equiv 2 + 23 \equiv 2 + 10 \equiv 12 \pmod{13}
   \]

3. Next, we compute \( 10^{2000} \mod 13 \):
   \[
   10^{2000} = (10^3)^{666} \cdot 10^2 \equiv (-1)^{666} \cdot 100 \equiv 1 \cdot 100 \equiv 100 \pmod{13}
   \]
   Since \( 100 \equiv 9 \pmod{13} \):
   \[
   10^{2000} \equiv 9 \pmod{13}
   \]

4. Now, we combine the results:
   \[
   N \equiv N' + x \cdot 10^{2000} \pmod{13}
   \]
   Substituting the values:
   \[
   0 \equiv 12 + 9x \pmod{13}
   \]
   Solving for \( x \):
   \[
   9x \equiv -12 \pmod{13}
   \]
   Since \( -12 \equiv 1 \pmod{13} \):
   \[
   9x \equiv 1 \pmod{13}
   \]
   To find \( x \), we need the multiplicative inverse of \( 9 \) modulo \( 13 \). The inverse of \( 9 \) modulo \( 13 \) is \( 3 \) because:
   \[
   9 \cdot 3 \equiv 27 \equiv 1 \pmod{13}
   \]
   Therefore:
   \[
   x \equiv 3 \pmod{13}
   \]

The final answer is \( \boxed{3} \)