55rd Ukrainian National Mathematical Olympiad - Fourth Round

It is obvious that the greater the number is, the more digits it has. Since every digit equals at least $1$, the maximum amount of digits is $2015$. Such number is unique: $\underline{11\dots 1}$, since every digit is $1$ and their sum is $2015$.

However, $111\dots 1$ (with $2015$ digits) is not divisible by $7$. For example, $11111=15873\cdot 7$, but $11111\underline{}7$ and $2015=335\cdot 6+5$, so this number is not divisible by $7$.

It is not possible that the number has $2014$ digits, since the sum of $2014$ odd numbers cannot be odd. That means that the maximum amount of digits in the number is $2013$. Obviously, it consists of $2012$ digits $1$ and one digit $3$, since no other cases are possible. We will find the maximum number with such structure that is divisible by $7$. That means we have to find such number with the digit $3$ in the maximum possible position. Thus the number will look like $A=\underline{11\dots 1}+2\cdot 10^k$, where $k\le 2012$, and $A$ is divisible by $7$ and $k$ is the maximum possible number.

Since $A=\overline{11\dots 1}+2\cdot 10^k$, and $2010=335\cdot 6$ and $11111:7$, then $B=111+2\cdot 10^k$. Since $111\equiv 6(\mathrm{mod}7)$, then $2\cdot 10^k\equiv 1(\mathrm{mod}7)$. Consider periodicity of remainders of such numbers modulo $7$ (remembering that $10\equiv 3(\mathrm{mod}7)$): $$\begin{gathered} 2\cdot 10^0=2(\mathrm{mod}7),2\cdot 10^1=6(\mathrm{mod}7),2\cdot 10^2=4(\mathrm{mod}7),2\cdot 10^3=5(\mathrm{mod}7), \\ 2\cdot 10^4=1(\mathrm{mod}7),2\cdot 10^5=3(\mathrm{mod}7),2\cdot 10^6=2(\mathrm{mod}7),\dots \end{gathered}$$ And so on. Thus values for $k$ will look like $k=6l+4$. Since $2010=335\cdot 6$, then the maximum possible value for $k=334\cdot 6+4=2008$, therefore, the maximum number is $11113\overline{11\dots 1}$ (with $2008$ digits $1$ after the $3$).