A6. The number $a=11 \ldots 111$ consists of exactly 2011 ones.

What is the remainder of $a$ when divided by 37?
A) 0
B) 1
C) 3
D) 7
E) 11
A6. B) 1 Door with a division of $a=11 \cdots 11$ (2011 ones) to be divided by 37, it quickly becomes apparent that 111 is divisible by 37. We will use this fact. We now see that the number $1110 \cdots 0$ is divisible by 37, regardless of the number of zeros at the end. In particular

![](https://cdn.mathpix.com/cropped/2024_04_17_7effb8aeb7220f8b3452g-2.jpg?height=49&width=1577&top_left_y=341&top_left_x=248)
$1110 \cdots 0$ (2002 zeros), and so on until 1110 (1 zero). The sum of these numbers is $1 \ldots 10$ (2010 ones) and is therefore also divisible by 37. The remainder of $a$ when divided by 37 is thus equal to 1, because $a-1$ is divisible by 37.