1. Find the numbers of the form $\overline{a b c d}$, written in base $10, \overline{c d}<\overline{a b}$, knowing that the product of the number formed by the first two digits of the number and the number formed by the last two digits of it is equal to 300, and $\frac{(\overline{a b}, \overline{c d})}{[\overline{a b}, \overline{c d}]}=0.08(3)$; where $(\overline{a b}, \overline{c d})$ and $[\overline{a b}, \overline{c d}]$ denote, respectively, the greatest common divisor and the least common multiple of the numbers $\overline{a b}$ and $\overline{c d}$.

Prof. Gheorghe Iacoviță
Solution: Let $\overline{ab}=m, \overline{cd}=n$. We can write $m>n, mn=300, \frac{(m, n)}{[m, n]}=0.08(3)$.

$\frac{(m, n)}{[m, n]}=0.08(3) \Leftrightarrow \frac{(m, n)}{[m, n]}=\frac{1}{12} \Leftrightarrow 12(m, n)=[m, n] ;$ from which $12(m, n)[m, n]=[m, n]^{2} ;$ and since $(m, m)[m, n]=mn,$ we have: $12 \cdot 300=[m, n]^{2} \Rightarrow [m, n]=60,$ thus $(m, n)=5$. Then $m=5x, n=5y, (x, y)=1$; in conclusion, we will have: $m \cdot n=300 \Rightarrow 5x \cdot 5y=300 \Rightarrow xy=12$.

Since $m>n \Rightarrow x>y$, we have the following cases:

1) $x=4, y=3 \Rightarrow m=20, n=15$; i.e., $\overline{ab}=20, \overline{cd}=15$, so $\overline{abcd}=2015$.
2) $x=6, y=2,$ but $(x, y) \neq 1$.
3) $x=12, y=1 \Rightarrow m=60, n=5$, which are not two-digit numbers.

In conclusion, $\overline{abcd}=2015$.

Scoring:

| Let $\overline{ab}=m, \overline{cd}=n \cdot \frac{(m, n)}{[m, n]}=0.08(3) \Leftrightarrow \frac{(m, n)}{[m, n]}=\frac{1}{12} \Leftrightarrow 12(m, n)=[m, n]$ | $\mathbf{1 p}$ |
| :--- | :--- |
| $12(m, m)[m, n]=[m, n]^{2}$ | $\mathbf{1 p}$ |
| Find $[m, n]=60$ and $(m, n)=5$ | $\mathbf{1 p}$ |
| $m=5x, n=5y, (x, y)=1 ;$ from which $m \cdot n=300 \Rightarrow 5x \cdot 5y=300 \Rightarrow xy=12$. | $\mathbf{1 p}$ |
| Discuss the cases. | $\mathbf{2 p}$ |
| Finalization. | $\mathbf{1 p}$ |