## Subject IV

Determine all pairs of natural numbers $\overline{a b}$ and $\overline{x y z}$, with $x<y<z$, such that

$$
\overline{a b} \cdot\left(x^{2}+y^{2}+z^{2}\right)=1 \cdot 2 \cdot 19 \cdot 53
$$

Note:

Working time: 2 hours

All subjects are mandatory

Each subject is graded from 0 to 7

No points are given ex officio

## MATHEMATICS OLYMPIAD

Local stage - Constanța, 23.02.2014

5th Grade

## Grading and marking rubric
## Subject IV

Determine all pairs of natural numbers $\overline{a b}$ and $\overline{x y z}$, with $x<y<z$, such that

$$
\overline{a b} \cdot\left(x^{2}+y^{2}+z^{2}\right)=1 \cdot 2 \cdot 19 \cdot 53
$$

## Solution

$\overline{a b} \in\{19,38,53\}$................................................................................................................................................

$\mathbf{1}^{0} . \overline{a b}=19 \Rightarrow x^{2}+y^{2}+z^{2}=106 \Rightarrow 3^{2}+4^{2}+9^{2}=106$, unique representation, so $\overline{x y z}=349 \ldots \ldots .2 \mathrm{p}$

$2^{0}$. $\overline{a b}=38 \Rightarrow x^{2}+y^{2}+z^{2}=53 \Rightarrow 1^{2}+4^{2}+6^{2}=53$, unique representation, so $\overline{x y z}=146 \ldots \ldots . . . .2 p$

$3^{0} . \overline{a b}=53 \Rightarrow x^{2}+y^{2}+z^{2}=38 \Rightarrow 2^{2}+3^{2}+5^{2}=38$, unique representation, so $\overline{x y z}=235$.

Therefore, the pairs $(\overline{a b} ; \overline{x y z})$ are: $(19; 349)$, $(38 ; 146)$, $(53 ; 235)$