8.1. Determine all natural numbers of the form $\overline{a b c d}$ that are divisible by 3 and simultaneously satisfy the conditions: $a+b+d=11, a+c+d=12, b+c+d=10$.
Solution:

1) Since $\overline{a b c d} \div 3$, we deduce that the sum of all digits is divisible by 3: $(a+b+c+d) \vdots 3$
2) By adding the three equalities from the statement, we obtain: $2 a+2 b+2 c+3 d=33 \Rightarrow$ $\Rightarrow 2(a+b+c)=33-3 d \Rightarrow 2(a+b+c)=3(11-d)$

From (1) we have $(a+b+c) \vdots 3$

3) From $a+b+d=11$ we get $a+b=11-d$

From (1) and (2) we get $2(a+b)+2 c=3(11-d) \Rightarrow 2(11-d)+2 c=3(11-d) \Rightarrow 2 c=11-d \Rightarrow$

$\Rightarrow c=\frac{11-d}{2} \quad$ (3) and thus $d$ is an odd digit.

But $(a+b+c) \vdots 3$ and $(a+b+c+d) \vdots 3 \Rightarrow d \vdots 3 \Rightarrow d \in\{3 ; 9\}$, since $d$ is an odd digit.

3) Let $d=3 \Rightarrow c=\frac{11-3}{2}=4 \Rightarrow c=4$.

But from the hypothesis we have $a+c+d=12 \Rightarrow a+4+3=12 \Rightarrow a=5$.

From $b+c+d=10$ (from the hypothesis) $\Rightarrow b+4+3=10 \Rightarrow b=3 \Rightarrow \overline{a b c d}=5343$

4) Let $d=9 \Rightarrow c=\frac{11-9}{2}=1 \Rightarrow c=1$.

But from the hypothesis we have $a+c+d=12 \Rightarrow a+9+1=12 \Rightarrow a=2$.

From $b+c+d=10$ (from the hypothesis) $\Rightarrow b+1+9=10 \Rightarrow b=0 \Rightarrow \overline{a b c d}=2019$

Answer: $\overline{a b c d} \in\{5343 ; 2019\}$.

## 63rd MATH OLYMPIAD OF THE REPUBLIC OF MOLDOVA  Chişinău, March 1, 2019 - March 4, 2019  First day, February 2, 2019. Grade 8

## Solution to problem 8.2