Frankin B.P.

Natural numbers $a<b<c$ are such that $b+a$ is divisible by $b-a$, and $c+b$ is divisible by $c-b$. The number $a$ is written with 2011 digits, and the number $b-2012$ is written with 2012 digits. How many digits does the number $c$ have?
According to the condition, the number $2 a=(b+a)-(b-a)$ is divisible by $b-a$. Therefore, $b-a \leq 2 a$, which means $b \leq 3 a$. Similarly, $c \leq 3b$. Thus, $c \leq 9 a < 10 a$, so in the notation with no more than 2012 digits (but not less, since $c \geq b$).

## Answer

2012 digits.