[b]p1.[/b] Suppose that we know that for all integer numbers $x$ the value of $ax^2 + bx + c$ is also integer. Can we conclude that the numbers $a, b, c$ are integers?


[b]p2.[/b] Let $x, y$ be integer numbers such that $3x + 7y$ is divisible by $19$. Prove that $43x + 75y$ is also divisible by $19$.


[b]p3.[/b] What is the biggest power of $2$ which divides the product of $2007$ consequitive numbers $2008 \cdot 2009 \cdot 2010 ... \cdot 4014$?


[b]p4.[/b] It is easy to show that the sum of five acute angles of a regular star is equal to $180$ degrees. Prove that the sum of five angles of an irregular star is also $180$ degrees.
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PS. You should use hide for answers.
**p1.** Suppose that we know that for all integer numbers \( x \) the value of \( ax^2 + bx + c \) is also an integer. Can we conclude that the numbers \( a, b, c \) are integers?

1. Consider the polynomial \( P(x) = ax^2 + bx + c \).
2. Since \( P(x) \) is an integer for all integer \( x \), we can evaluate \( P(x) \) at specific integer values to derive conditions on \( a, b, \) and \( c \).
3. Evaluate \( P(0) \):
   \[
   P(0) = c
   \]
   Since \( P(0) \) is an integer, \( c \) must be an integer.
4. Evaluate \( P(1) \):
   \[
   P(1) = a + b + c
   \]
   Since \( P(1) \) is an integer and \( c \) is already an integer, \( a + b \) must be an integer.
5. Evaluate \( P(2) \):
   \[
   P(2) = 4a + 2b + c
   \]
   Since \( P(2) \) is an integer and \( c \) is already an integer, \( 4a + 2b \) must be an integer.
6. From the conditions \( a + b \) and \( 4a + 2b \) being integers, we can solve for \( a \) and \( b \):
   \[
   4a + 2b - 2(a + b) = 2a
   \]
   Since \( 4a + 2b \) and \( a + b \) are integers, \( 2a \) must be an integer, implying \( a \) is an integer.
7. Since \( a \) is an integer, \( a + b \) being an integer implies \( b \) must also be an integer.

Thus, \( a, b, \) and \( c \) are all integers.

\(\blacksquare\)

**p2.** Let \( x, y \) be integer numbers such that \( 3x + 7y \) is divisible by \( 19 \). Prove that \( 43x + 75y \) is also divisible by \( 19 \).

1. Given \( 3x + 7y \equiv 0 \pmod{19} \), we need to show \( 43x + 75y \equiv 0 \pmod{19} \).
2. Reduce \( 43 \) and \( 75 \) modulo \( 19 \):
   \[
   43 \equiv 5 \pmod{19} \quad \text{and} \quad 75 \equiv 18 \equiv -1 \pmod{19}
   \]
3. Substitute these into the expression:
   \[
   43x + 75y \equiv 5x - y \pmod{19}
   \]
4. We need to show \( 5x - y \equiv 0 \pmod{19} \).
5. From \( 3x + 7y \equiv 0 \pmod{19} \), solve for \( y \):
   \[
   7y \equiv -3x \pmod{19} \implies y \equiv -3x \cdot 7^{-1} \pmod{19}
   \]
6. Find the modular inverse of \( 7 \) modulo \( 19 \):
   \[
   7 \cdot 3 = 21 \equiv 2 \pmod{19} \implies 7^{-1} \equiv 11 \pmod{19}
   \]
7. Substitute back:
   \[
   y \equiv -3x \cdot 11 \equiv -33x \equiv 5x \pmod{19}
   \]
8. Substitute \( y \equiv 5x \) into \( 5x - y \):
   \[
   5x - y \equiv 5x - 5x \equiv 0 \pmod{19}
   \]

Thus, \( 43x + 75y \) is divisible by \( 19 \).

\(\blacksquare\)

**p3.** What is the biggest power of \( 2 \) which divides the product of \( 2007 \) consecutive numbers \( 2008 \cdot 2009 \cdot 2010 \cdots 4014 \)?

1. The given product is equal to \( \frac{4014!}{2007!} \).
2. By Legendre's formula, the highest power of a prime \( p \) dividing \( n! \) is given by:
   \[
   \nu_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor
   \]
3. Calculate \( \nu_2(4014!) \):
   \[
   \nu_2(4014!) = \left\lfloor \frac{4014}{2} \right\rfloor + \left\lfloor \frac{4014}{4} \right\rfloor + \left\lfloor \frac{4014}{8} \right\rfloor + \left\lfloor \frac{4014}{16} \right\rfloor + \left\lfloor \frac{4014}{32} \right\rfloor + \left\lfloor \frac{4014}{64} \right\rfloor + \left\lfloor \frac{4014}{128} \right\rfloor + \left\lfloor \frac{4014}{256} \right\rfloor + \left\lfloor \frac{4014}{512} \right\rfloor + \left\lfloor \frac{4014}{1024} \right\rfloor + \left\lfloor \frac{4014}{2048} \right\rfloor
   \]
   \[
   = 2007 + 1003 + 501 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 4005
   \]
4. Calculate \( \nu_2(2007!) \):
   \[
   \nu_2(2007!) = \left\lfloor \frac{2007}{2} \right\rfloor + \left\lfloor \frac{2007}{4} \right\rfloor + \left\lfloor \frac{2007}{8} \right\rfloor + \left\lfloor \frac{2007}{16} \right\rfloor + \left\lfloor \frac{2007}{32} \right\rfloor + \left\lfloor \frac{2007}{64} \right\rfloor + \left\lfloor \frac{2007}{128} \right\rfloor + \left\lfloor \frac{2007}{256} \right\rfloor + \left\lfloor \frac{2007}{512} \right\rfloor + \left\lfloor \frac{2007}{1024} \right\rfloor
   \]
   \[
   = 1003 + 501 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 1998
   \]
5. The highest power of \( 2 \) dividing \( \frac{4014!}{2007!} \) is:
   \[
   \nu_2(4014!) - \nu_2(2007!) = 4005 - 1998 = 2007
   \]

\(\boxed{2007}\)

**p4.** It is easy to show that the sum of five acute angles of a regular star is equal to \( 180 \) degrees. Prove that the sum of five angles of an irregular star is also \( 180 \) degrees.

1. Consider a star polygon with five vertices and five angles at the vertices.
2. Each vertex angle is an external angle of the star.
3. The sum of the external angles of any polygon is \( 360 \) degrees.
4. In a star polygon, each external angle is counted twice, once for each of the two intersecting lines forming the star.
5. Therefore, the sum of the angles at the vertices of the star is:
   \[
   \frac{360}{2} = 180 \text{ degrees}
   \]

\(\blacksquare\)