Anurag@Gurome wrote:

gmatapril wrote:If n is a prime number greater than 3, what is the
remainder when n^2 is divided by 12 ?

....

my doubts are
1. why they have used no. 6 to divide by n why not 12
2.why the remainder cannot be 2 or 4 how did we know that n will be even then.

This problem can be easily solved if you know a property of any prime number greater than 3. In fact, the solution also uses the fact, but indirectly. The property is : Any prime number greater than 3 can be expressed as either (6a + 1) or (6a - 1), where a is a positive integer. There is a simple logic behind this. Let me explain why...

if we were asked any prime number greater than 7 how will we express it in terms of a

gmatapril wrote:If n is a prime number greater than 3, what is the
remainder when n^2 is divided by 12 ?
Answer is 1

According to OG explanation
For the more mathematically inclined, consider
the remainder when each prime number n greater
than 3 is divided by 6. Th e remainder cannot be 0
because that would imply that n is divisible by 6,
which is impossible since n is a prime number.
Th e remainder cannot be 2 or 4 because that
would imply that n is even, which is impossible
since n is a prime number greater than 3. Th e
remainder cannot be 3 because that would imply
that n is divisible by 3, which is impossible since
n is a prime number greater than 3. Th erefore,
the only possible remainders when a prime
number n greater than 3 is divided by 6 are 1
and 5. Th us, n has the form 6q + 1 or 6q + 5,
where q is an integer, and, therefore, n2 has the
form 36q2 + 12q + 1 = 12(3q2 + q) + 1 or
36q2 + 60q + 25 = 12(3q2 + 5q + 2) + 1. In either
case, n^2 has a remainder of 1 when divided by 12.



my doubts are
1. why they have used no. 6 to divide by n why not 12
2.why the remainder cannot be 2 or 4 how did we know that n will be even then.

please clear my doubts