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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