3. The fraction $\frac{2 x+3}{2 x-7}$ is given, with $x \in \mathbb{N}, x \geq 4$. Let $x_{1}, x_{2}, \ldots, x_{1000}$ be the first 1000 natural numbers for which the fraction can be simplified.

a) Calculate $x_{1000}$.

b) Show that $x_{p}+x_{q}-x_{p+q}=1$ for any $p, q \in\{1,2, \ldots, 1000\}$.

## Note: All subjects are mandatory

Each subject is graded from 0 to 7 points

Working time 2 hours

The problems were proposed and selected by:

Conf.univ.Horvat Marc Andrei- North University Center Baia Mare

Prof. Darolți Erica- National College "V.Lucaciu" Baia Mare

Prof. Serasz Maria- Gymnasium School "D. Cantemir" Baia Mare

BAREME - 6th grade
## Solution. 1 .

a) $2+4+6+\cdots+4032=2016 \cdot 2017$

1 point

$\frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\cdots+\frac{2}{2015 \cdot 2017}=\frac{2016}{2017}$

2 points

$N=2016^{2}$ perfect square

1 point

b) Let $x=m(\widehat{C O D})$. Then $x=\frac{2}{5}(180-2 x)$

1 point

$x=40^{\circ}$

1 point

Let $[O M$ be the bisector of $\widehat{C O D}$ and $[O N$ be the bisector of $\widehat{B O D}$

$m(\widehat{M O D})=20^{\circ}, m(\widehat{D O N})=50^{\circ}$

$m(\widehat{M O N})=70^{\circ}$

. .1 point

## Solution. 2.

a) $A_{9} A_{10}=18 \mathrm{~cm}, A_{9} M=M A_{10}=9 \mathrm{~cm}$
0.5 points
$A_{24} A_{25}=48 \mathrm{~cm}, A_{24} N=N A_{25}=24 \mathrm{~cm}$
0.5 points
$M N=M A_{10}+A_{10} A_{11}+\cdots+A_{23} A_{24}+A_{24} N$
1 point
$M N=9+(20+22+\cdots+46)+24$
1 point
$M N=495 \mathrm{~cm}$
1 point
b) $A_{1} A_{n}=x+2 x+\cdots+(n-1) x=x \cdot \frac{(n-1) n}{2}$
1 point
$n(n-1) \cdot x=2^{2} \cdot 3 \cdot 5 \cdot 29$
0.5 points
$n(n-1) \mid 2^{2} \cdot 3 \cdot 5 \cdot 29 \Rightarrow n \in\{2,3,4,5,6,30\}$
0.5 points

$$
\begin{gathered}
n=2 \Rightarrow x=870 \\
n=3 \Rightarrow x=290 \\
n=4 \Rightarrow x=145 \\
n=5 \Rightarrow x=87 \\
n=6 \Rightarrow x=58 \\
n=30 \Rightarrow x=2
\end{gathered}
$$

Solution. 3.

a) The fraction $\frac{2 x+3}{2 x-7}$ is reducible if there exists $d \in \mathbb{N}^{*}, d \neq 1$ such that $d \mid 2 x+3$ and $d \mid 2 x-7 \quad \mathbf{1}$ point Since $d \in\{2,5,10\}$ and $2 x+3,2 x-7$ are odd numbers, it follows that $d=5$

1 point

$2 x+3=5 k, k$ is an odd number

1 point

$k=2 p+1, x_{p}=5 p+1, p \geq 1$

1 point

$x_{1000}=5001$

1 point

b) $x_{p}=5 p+1, x_{q}=5 q+1, x_{p+q}=5(p+q)+1$

1 point

Verify the relation

1 point