## Subject I. (20 points)

Determine the number of elements of the set: $A=\left\{\overline{a b c d} \left\lvert\, \frac{a+3}{b}=\frac{b-3}{c}=\frac{c+4}{d}=\frac{d-4}{a}\right.\right\}$.

Prof. Ioan Balica, Ioan Bob Gymnasium School, Cluj-Napoca
## Subject I.

According to the property of the sequence of equal ratios, we have $\frac{a+3}{b}=\frac{b-3}{c}=\frac{c+4}{d}=\frac{d-4}{a}=\frac{a+3+b-3+c+4+d-4}{a+b+c+d}=1$.

Thus, $a+3=b$, $b-3=c$, and $c+4=d$. We obtain $c=a$ and $d=a+4$. Since $a, b, c, d$ are digits, $a \neq 0$. It follows that $a \in\{1,2,3,4,5\}$. We get $A=\{1415,2526,3637,4748,5859\}$, so $A$ has 5 elements.