## Subject II (7 points)

Determine the digits $\mathrm{a}$ and $\mathrm{b}$ (in base 10) knowing that the rational number $\quad r=\overline{a, 2(b)}+\overline{b, 3(a)}$ can be written as a finite decimal fraction.
## Solution:

We have $\mathrm{r}=\overline{a, 2(b)}+\overline{b, 3(a)}=\frac{\overline{a 2 b}-\overline{a 2}}{90}+\frac{\overline{b 3 a}-\overline{b 3}}{90}=\frac{90 \cdot a+b+18}{90}+\frac{a+90 \cdot b+27}{90}=\frac{91 \cdot a+91 \cdot b+45}{90}$ (2p)

(1p)

But $\mathrm{r}$ is a finite decimal fraction only if 9 divides $91 \cdot a+91 \cdot b+45$ $9 \mid (91(a+b)+45) \Rightarrow 9 \mid (91(a+b)) \Rightarrow 9 \mid (a+b)$

But a, b $\in\{0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8\} \Rightarrow 0 \leq \mathrm{a}+\mathrm{b} \leq 16 \Rightarrow \mathrm{a}+\mathrm{b}=0$ or a+b=9

$\Rightarrow($ a;b) $\in\{(0 ; 0) ;(1 ; 8) ;(2 ; 7) ;(3 ; 6) ;(4 ; 5) ;(5 ; 4) ;(6 ; 3) ;(7 ; 2) ;(8 ; 1)\}$