The first student attempted all the questions i.e. question number 1, 2, 3, ..., 999, 1000.
The second student attempted the questions with numbers which were multiple of 2 i.e. question number 2, 4, 6, ..., 998, 1000.
The third student attempted the questions with numbers which were multiple of 3 i.e. question number 3, 6, 9, 12, ..., 999.
And so on.
Here we can see that the number of students who have attempted a particular question is the number of factors (including the number itself and 1) it has. For example, f(24) = 8 as the question number 24 was attempted by 8 students (1^{st}, 2^{nd}, 3^{rd}, 4^{th}, 6^{th}, 8^{th}, 12^{th} and 24^{th} student).
∴ We have to find the question numbers for which the number of factors are odd.
We know that any number 'n' can be expressed in the form,
n = 2^{(b)}3^{(c)}5^{(d)}... where 2, 3, 5,.... are prime numbers
The number of factors for n = (b + 1)(c + 1)(d + 1)...
To have odd number of factors each of (b + 1), (c + 1), (d + 1), ... should be odd.
∴ b, c, d, ... all should be even.
∴ n is a ^{}^{}^{}perfect square.
∴ The questions with numbers which were perfect squares were attempted by odd number of students.
∴ The problem reduces to finding out the number of perfect squares in the range 1  1000.
There are 31 perfect squares between 1 to 1000 including 1 and 1000.
Hence, option 2.
