I would go for C although I'm not sure of an algebraic proof? Is there one?

I tried to solve this question by picking numbers.

I. lets say number - 5 (n= 1) 25 (n= 2)
- 13 (n=2) 169 (n=3) => (2n-1 digits)
Satisfies

II. lets consider the same numbers. 25 can be written as 3*8 +1
Similarly even 169 can be written in the form of 3n +1. You can check for any square number, you will see it satisfies this criteria.

III. say 25 can be written as 4*6 + 1
similarly you can check for other square numbers also. You will find it satisfies the criteria.

So my take is E

I think I agree with you

II. would fail if n =2,
3n+1 = 7 which is not a square number

III. would fail if n = 1,
4n+1 = 5 which is again not a square number

I. Is always true

I'm not sure if I did something wrong.. but I tested with 5 and II and III were proven incorrect.

note the term 'n' in I is different from II and III
In I, 'n' denotes the number of digits, which is not the case in II and III.
Also in II, we are talking about squared numbers. 5 is not a square number. 25 is a square of number 5, so we can call 25 to be square number

Hope this helps!

yeah.. I see it.... thanks!

Note that each integer can be written as 2k or 2k + 1


Also, each integer can be written as 3k - 1 , 3k, or 3k + 1