All digits, except the 44th digit (from the left), of an 80-digit posi

: If we have a 6-digit number with all digits identical, say dddddd, where d is a digit, this number is divisible by 7, 11 and 13. Why?
Note: dddddd = ddd x 1000 + ddd = ddd x (1000 + 1)
= ddd x 1001
= d x (111) x (1001)
= d x (3 x 37) x (7 x 11 x 13)

Now, we have 80 digit number where all digits are 2 except the 44th digit from the left.

Let us group six 2s at a time starting from the left (shown below):

222222222222 ...
---6---|---6---|

We will get 7 groups and reach the 42nd digit
Next, we have the 43rd digit which is also 2; and the 44th digit - say d.

After this, there are 36 more digits (all 2s), which can also be grouped six at a time to form 6 groups
The 7 groups initially and the 6 groups at the end are all divisible by 13

Thus, we just need the number formed by the 42nd and the 43rd digits, i.e. "2x" to be divisible by 13
Thus, x = 6 (since 26 is divisible by 13)
correct answer E