Correct option is (D) 1
Any prime number greater than 3 is of the form \(6k\pm1,\) where k is a natural number.
Thus, \((6k\pm1)^2\) \(=36k^2\pm12k+1\)
= 6k (6k \(\pm\) 2) + 1
When \((6k\pm1)^2\) or (6k (6k \(\pm\) 2) + 1) is divided by 6, we get k (6k \(\pm\) 2) as a quotient and 1 as a remainder.
\(\therefore\) When the square of any prime number greater than 3 is divided by 6, we get 1 as a remainder.
