Question

What is the remainder when $p^2+17$ is divided by 12, where 'p' is a four digit prime number?

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

The correct option is D 6

Prime number greater than 3 can be written in the form of 6N + 1 or 6N - 1
So, $P=(6N\pm 1{)}^2+17$
$=36N^2\pm 12N+18$
$=36N^2\pm 12N+12+6$
$=12(3N^2\pm N)+6$
Hence, remainder = 6