Problem 1658

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: $\underline{11\dots1}$, since every digit is $1$ and their sum is $2015$.

However, $111\dots1$ (with $2015$ digits) is not divisible by $7$. For example, $11111 = 15873 \cdot 7$, but $11111\underline{\hspace{2cm}}7$ 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 $A = \underline{11\dots1} + 2 \cdot 10^k$, where $k \le 2012$, and $A$ is divisible by $7$ and $k$ is the maximum possible number.

Since $A = \overline{11\dots1} + 2 \cdot 10^k$, and $2010 = 335 \cdot 6$ and $11111:7$, then $B = 111 + 2 \cdot 10^k$. Since $111 \equiv 6 \pmod{7}$, then $2 \cdot 10^k \equiv 1 \pmod{7}$. Consider periodicity of remainders of such numbers modulo $7$ (remembering that $10 \equiv 3 \pmod{7}$):
$$2 \cdot 10^0 = 2 \pmod{7},\quad 2 \cdot 10^1 = 6 \pmod{7},\quad 2 \cdot 10^2 = 4 \pmod{7},\quad 2 \cdot 10^3 = 5 \pmod{7},\\ 2 \cdot 10^4 = 1 \pmod{7},\quad 2 \cdot 10^5 = 3 \pmod{7},\quad 2 \cdot 10^6 = 2 \pmod{7}, \dots$$
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 $11113\overline{11\dots1}$ (with $2008$ digits $1$ after the $3$).