p1. Let $A = \{(x, y)|3x + 5y\ge 15, x + y^2\le 25, x\ge 0, x, y$ integer numbers $\}$. Find all pairs of $(x, zx)\in A$ provided that $z$ is non-zero integer.


p2. A shop owner wants to be able to weigh various kinds of weight objects (in natural numbers) with only $4$ different weights.
(For example, if he has weights $ 1$, $2$, $5$ and $10$. He can weighing $ 1$ kg, $2$ kg, $3$ kg $(1 + 2)$, $44$ kg $(5 - 1)$, $5$ kg, $6$ kg, $7$ kg, $ 8$ kg, $9$ kg $(10 - 1)$, $10$ kg, $11$ kg, $12$ kg, $13$ kg $(10 + 1 + 2)$, $14$ kg $(10 + 5 -1)$, $15$ kg, $16$ kg, $17$ kg and $18$ kg). If he wants to be able to weigh all the weight from $ 1$ kg to $40$ kg, determine the four weights that he must have. Explain that your answer is correct.


p3. Given the following table.
[img]https://cdn.artofproblemsolving.com/attachments/d/8/4622407a72656efe77ccaf02cf353ef1bcfa28.png[/img]
Table $4\times 4$ ​​is a combination of four smaller table sections of size $2\times 2$. 
This table will be filled with four consecutive integers such that:
$\bullet$ The horizontal sum of the numbers in each row is $10$ .
$\bullet$  The vertical sum of the numbers in each column is $10$
$\bullet$  The sum of the four numbers in each part of $2\times 2$ which is delimited by the line thickness is also equal to $10$.
Determine how many arrangements are possible.


p4. A sequence of real numbers is defined as following:
$U_n=ar^{n-1}$, if $n = 4m -3$ or $n = 4m - 2$
$U_n=- ar^{n-1}$,  if $n = 4m - 1$ or $n = 4m$,  where $a > 0$, $r > 0$, and $m$ is a  positive integer.
Prove that the sum of all the $ 1$st to $2009$th terms is $\frac{a(1+r-r^{2009}+r^{2010})}{1+r^2}$


5. Cube $ABCD.EFGH$ is cut into four parts by two planes. The first plane is parallel to side $ABCD$ and passes through the midpoint of edge $BF$. The sceond plane passes through the midpoints $AB$, $AD$, $GH$, and $FG$. Determine the ratio of the volumes of the smallest part to the largest part.
To prove that the sum of the first 2009 terms of the sequence \( U_n \) is given by \(\frac{a(1+r-r^{2009}+r^{2010})}{1+r^2}\), we will follow these steps:

1. **Redefine the sequence \( U_n \) based on the given conditions:**
   \[
   U_n = \begin{cases} 
   ar^{n-1} & \text{if } n = 4m - 3 \text{ or } n = 4m - 2, \\
   -ar^{n-1} & \text{if } n = 4m - 1 \text{ or } n = 4m 
   \end{cases}
   \]
   where \( m \) is a positive integer.

2. **Group the terms in sets of four:**
   Notice that every four terms form a pattern:
   \[
   (ar^{4m-4}, ar^{4m-3}, -ar^{4m-2}, -ar^{4m-1})
   \]
   For example, the first four terms are:
   \[
   (a, ar, -ar^2, -ar^3)
   \]
   The next four terms are:
   \[
   (ar^4, ar^5, -ar^6, -ar^7)
   \]
   And so on.

3. **Sum the terms within each group of four:**
   The sum of each group of four terms is:
   \[
   ar^{4m-4} + ar^{4m-3} - ar^{4m-2} - ar^{4m-1}
   \]
   Factor out the common term \( ar^{4m-4} \):
   \[
   ar^{4m-4}(1 + r - r^2 - r^3)
   \]

4. **Sum the series up to the 2008th term:**
   Since 2009 is not a multiple of 4, we need to handle the last term separately. The sum of the first 2008 terms can be written as:
   \[
   S = \sum_{m=1}^{502} ar^{4m-4}(1 + r - r^2 - r^3)
   \]
   This is a geometric series with the common ratio \( r^4 \):
   \[
   S = (1 + r - r^2 - r^3) \sum_{m=0}^{501} ar^{4m}
   \]
   The sum of the geometric series is:
   \[
   \sum_{m=0}^{501} ar^{4m} = a \frac{1 - (r^4)^{502}}{1 - r^4}
   \]
   Therefore:
   \[
   S = a(1 + r - r^2 - r^3) \frac{1 - r^{2008}}{1 - r^4}
   \]

5. **Add the 2009th term:**
   The 2009th term is \( ar^{2008} \). So, the total sum \( S_{total} \) is:
   \[
   S_{total} = a(1 + r - r^2 - r^3) \frac{1 - r^{2008}}{1 - r^4} + ar^{2008}
   \]

6. **Simplify the expression:**
   Combine the terms:
   \[
   S_{total} = a \left( \frac{(1 + r - r^2 - r^3)(1 - r^{2008})}{1 - r^4} + r^{2008} \right)
   \]
   Simplify the numerator:
   \[
   (1 + r - r^2 - r^3)(1 - r^{2008}) = 1 + r - r^2 - r^3 - r^{2008} - r^{2009} + r^{2010} + r^{2011}
   \]
   Therefore:
   \[
   S_{total} = a \left( \frac{1 + r - r^2 - r^3 - r^{2008} - r^{2009} + r^{2010} + r^{2011}}{1 - r^4} + r^{2008} \right)
   \]
   Combine the terms:
   \[
   S_{total} = a \left( \frac{1 + r - r^{2009} + r^{2010}}{1 + r^2} \right)
   \]

Thus, the sum of the first 2009 terms is:
\[
\boxed{\frac{a(1 + r - r^{2009} + r^{2010})}{1 + r^2}}
\]