scthakur
It will be C.
Try with numbers.
From stmt1:
n = 1,3,5,7,9,11,.....
Hence, remainder of (n-1)(n+1) will be different for different values of n....insufficient.
From stmt2:
n = 1,2,4,5,7,8,10,..... and again the desired remainder will be different for different values of n.
Combining stmt1 and stmt2:
n = 1, 5, 7, 11, ......and remainder will always be 0 for any value of n.....sufficient.
I don't think you can simply calculate some first prime number to get the conclusion
Among 3 consecutive integer n - 1, n, n + 1
n is not factor of 3 => n - 1 or n + 1 is divisible by 3
=> (n-1)(n+1) divisible by 3 (1)
n is not factor of 2 => n - 1 or n + 1 is divisible by 2
between n - 1 and n + 1, if n - 1 is divisible by 2, n + 1 is divisible by 4, if n - 1 is divisible by 4, n + 1 is divisible by 2
=> (n-1)(n+1) is divisible by 8. (2)
from (1) & (2), (n-1)(n+1) is divisible by 12, r = 0