## SUBJECT I (7p)

Determine the non-zero natural numbers a and b with the properties: $a+b=2014$ and $[a, b]=(a, b) \cdot 682$, where [a,b] is the least common multiple of the numbers a and b, and ( $a, b$ ) is the greatest common divisor of the numbers a and b.
## Subject I

Let $d=(a, b)$; then there exist non-zero natural numbers $p$ and $q$ such that $a=p d, b=q d$ and $(p, q)=1$.

We obtain $[a, b]=$ pqd and the properties from the hypothesis become:

$p d+q d=2014$ and $p q d=682 d \Rightarrow p q=682$ and $(p+q)$ divides $2014, 2014=2 \cdot 19 \cdot 53 . \quad 1 p$

But $682=2 \cdot 11 \cdot 31=\mathrm{pq},(\mathrm{p}, \mathrm{q})=1$.

$1 \mathrm{p}$

If $\mathrm{p}=2 \Rightarrow \mathrm{q}=341 \Rightarrow \mathrm{p}+\mathrm{q}=343=7^{3}$, which does not divide 2014.

If $\mathrm{p}=11 \Rightarrow \mathrm{q}=62 \Rightarrow \mathrm{p}+\mathrm{q}=73$, which does not divide 2014.

If $p=31 \Rightarrow q=22 \Rightarrow p+q=53$, which divides 2014.

Since the other possible cases reduce to these, by swapping $a$ with $b$, the only suitable situation is $p=31, q=22$, from which $d=2014:(p+q)=2014: 53=38$.

We obtain $a=31 \cdot 38=1178$ and $b=22 \cdot 38=836$.