What is the remainder when the square of any prime number greater than 3 is divided by 12?

The remainder when the square of any prime number greater than 3 is divided by 6, is

Any prime number greater than 3 is of the form `6k +- 1`, where k is a natural number.

Thus,

`( 6k +- 1)^2= 36k^2+-12k+1`

`= 6k (6k+-2)+1`

When,  `6k(6k+-2)+1` is divided by 6, we get, `k(6k+-2)`and remainder as 1.

Hence, the correct choice is (a).

[Hint: Any prime number greater than 3 is of the from 6k ± 1, where k is a natural number and (6k ± 1)2 = 36k2 ± 12k + 1 = 6k(6k ± 2) + 1]

Concept: Euclid’s Division Lemma

  Is there an error in this question or solution?

The remainder when the square of any prime number greater than 3 is divided by 6, is

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