p1. Known points $A (-1.2)$, $B (0,2)$, $C (3,0)$, and $D (3, -1)$ as seen in the following picture.
 Determine the measure of the angle $AOD$ .
[img]https://cdn.artofproblemsolving.com/attachments/f/2/ca857aaf54c803db34d8d52505ef9a80e7130f.png[/img]


p2. Determine all prime numbers $p> 2$ until $p$ divides $71^2 - 37^2 - 51$.


p3. A ball if dropped perpendicular to the ground from a height then it will bounce back perpendicular along the high third again, down back upright and bouncing back a third of its height, and next. If the distance traveled by the ball when it touches the ground the fourth time is equal to $106$ meters. From what height is the ball was dropped?


p4. The beam $ABCD.EFGH$ is obtained by pasting two unit cubes $ABCD.PQRS$ and $PQRS.EFGH$. The point K is the midpoint of the edge $AB$, while the point $L$ is the midpoint of the edge $SH$. What is the length of the line segment $KL$?


p5. How many integer numbers are no greater than $2004$, with  remainder $1$ when divided by $2$, with  remainder $2$ when divided by $3$, with  remainder $3$ when divided by $4$, and with  remainder $4$ when divided by $5$?
### Problem 1:
To determine the measure of the angle \( \angle AOD \), we need to use the coordinates of points \( A \), \( O \), and \( D \).

1. **Identify the coordinates:**
   - \( A(-1, 2) \)
   - \( O(0, 0) \)
   - \( D(3, -1) \)

2. **Find the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OD} \):**
   \[
   \overrightarrow{OA} = (-1, 2)
   \]
   \[
   \overrightarrow{OD} = (3, -1)
   \]

3. **Use the dot product to find the cosine of the angle \( \theta \) between the vectors:**
   \[
   \overrightarrow{OA} \cdot \overrightarrow{OD} = (-1) \cdot 3 + 2 \cdot (-1) = -3 - 2 = -5
   \]
   \[
   \|\overrightarrow{OA}\| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
   \]
   \[
   \|\overrightarrow{OD}\| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}
   \]
   \[
   \cos \theta = \frac{\overrightarrow{OA} \cdot \overrightarrow{OD}}{\|\overrightarrow{OA}\| \|\overrightarrow{OD}\|} = \frac{-5}{\sqrt{5} \cdot \sqrt{10}} = \frac{-5}{\sqrt{50}} = \frac{-5}{5\sqrt{2}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}
   \]

4. **Find the angle \( \theta \):**
   \[
   \theta = \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = 135^\circ
   \]

The final answer is \( \boxed{ 34 } \)