## SUBJECT 1

The sum of two natural numbers is 2016. If both numbers are divided by 4, the difference between the quotients is 468. Find the numbers.
## SUBJECT 1

$$
\begin{aligned}
& a=4 \cdot c_{1}+r_{1}, r_{1}<4 \\
& b=4 \cdot c_{2}+r_{2}, r_{2}<4
\end{aligned}
$$

$4 \cdot\left(c_{1}+c_{2}\right)+r_{1}+r_{2}=2016 \in M_{4} \Rightarrow r_{1}+r_{2} \in M_{4}$ ..... $1 p$
$r_{1}, r_{2} \in\{0 ; 1 ; 2 ; 3\} \Rightarrow r_{1}+r_{2} \in\{0 ; 4\}$ ..... 1p
$2 p$
$r_{1}+r_{2}=4 \Rightarrow c_{1}+c_{2}=503$ and $c_{1}-c_{2}=468 \Rightarrow c_{1} \notin \mathbb{N}$ ..... $1 p$
$\mathrm{a}=4 \cdot 486=1944$ and $b=4 \cdot 18=72$ ..... $1 p$