Baltic Way 2023 Shortlist

The answer is 2137.

First of all, let's observe that $x=63$, $y=2137$ satisfies the given relation. In fact, $$\begin{gathered} \frac{S(63)+S(1)}{S(64)}=\frac{77+1}{100}=\frac{78}{100}<\frac{8}{10}\text{and} \\ \frac{S(2137)+S(2023)}{S(4160)}=\frac{4131+3747}{10100}=\frac{7878}{10100}=\frac{78}{100}<\frac{8}{10}. \end{gathered}$$

Now let's assume that the octal representation of $x$ ends with exactly $k$ digits “7” (maybe $k=0$). In other words, $x=a\cdot 8^k+8^k-1$ where $a$ is a nonnegative integer satisfying $a\mathrm{mod}8<7$. In particular, we know that $S(a+1)=S(a)+1$ because there is no carry in addition in the octal system, since $a\mathrm{mod}8<7$. Therefore, $$\begin{gathered} S(x)=S(a\cdot 8^k+8^k-1)=S(a)\cdot 10^k+\frac{7}{9}(10^k-1)\text{and} \\ S(x+1)=S((a+1)\cdot 8^k)=S(a+1)\cdot 10^k=(S(a)+1)\cdot 10^k. \end{gathered}$$

Let's observe that $k=0$ would imply that $S(x)+S(1)=S(x+1)$ and therefore $k\ne 0$. $k\ge 3$ would imply $x\ge 8^3-1>100$ and therefore $k=1$ or $k=2$. If $k=2$ then $a=0$ (because $8^2+8^2-1>100$) and therefore $x=8^2-1=63$ and $\frac{S(x)+S(1)}{S(x+1)}=\frac{78}{100}$. $k=1$ is also impossible because: $$\frac{10S(a)+8}{10S(a)+10}<\frac{8}{10}⟹10S(a)+8<8S(a)+8⟹S(a)<0.$$

Now let's assume that we can find a solution with $y<2137=4131_8$. Then $y$ has at most 4 digits in its octal representation. Let $d_i$ for $i=0,1,2,3$ be equal to 1 if there is a carry at the $i$-th position in addition of $y$ and $2023=3747_8$ in octal system. It is easy to conclude that $$S(y+2023)-S(y)-S(2023)=\sum _{i=0}^3(-8)\cdot 10^i\cdot d_i+\sum _{i=0}^{3}10^{i+1}\cdot d_i=2\sum _{i=0}^{3}10^i\cdot d_i.$$ We have the implication: $$\begin{array}{rlrlr}\frac{S(y)+S(2023)}{S(y+2023)}& =\frac{78}{100}& ⟹& \frac{S(y)+S(2023)}{S(y+2023)-S(y)-S(2023)}& =\frac{78}{22}\\ & ⟹& S(y)+S(2023)& =\frac{78}{11}\sum _{i=0}^{3}10^i\cdot d_i.& \end{array}$$ We know that $d_3=1$ because otherwise we would have: $$S(y)+S(2023)\le \frac{78}{11}\cdot 111<3747=S(2023).$$ Because 78 is coprime with 11, we need to have the divisibility $11|{\sum }_{i=0}^{3}10^i\cdot d_i$. Since $d_3=1$, there are only three possible cases: ${\sum }_{i=0}^{3}10^i\cdot d_i=1001,1100,1111$. They correspond to $S(y)+S(2023)=7098,7800,7878$, so $S(y)=3351,4053,4131$. The first two cases are impossible because they assume $d_1=0$ but $5+4>7$, so there has to be a carry at the first position.