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

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

(a) 1

(b) 3

(c) 2

(d) 4

[Hint: Any prime number greater than 3 is of the from $6 k \pm 1$, where $k$ is a natural number and $(6 k \pm 1)^{2}=36 k^{2} \pm 12 k+$ $1=6 k(6 k \pm 2)+1]$

Solution:

Any prime number greater than 3 is of the form $6 k \pm 1$, where $k$ is a natural number.

Thus,

$(6 k \pm 1)^{2}=36 k^{2} \pm 12 k+1$

$=6 k(6 k \pm 2)+1$

When, $6 k(6 k \pm 2)+1$ is divided by 6 , we get, $k(6 k \pm 2)$ and remainder as $1 .$

Hence, the correct choice is (a).

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