Algebric/conceptual approach:


What is the remainder when n^2-1 / 8
or
What is the remainder when (n-1) (n+1) / 8

Stmt I

n is odd

When n is odd both n-1 and n+1 are 2 consecutive even integers. The product of any 2 consecutive even integers is always disivible by 8. The remainder is 0

2*4
4*6 ... so on

(Also note one of the 2 consecutive even integer is always divisible by 4 in a product of 2 consecutive even integers)

SUFF


Stmt II

n is not divisible by 8

n=2

(n-1) (n+1) / 8 = (1*3)/8 remainder= 3

n=5

(n-1) (n+1) / 8 = (4*6)/8 remainder= 0

INSUFF(many different values possible two of which are listed above)

Choose A


Hope this helps!

Regards,
CR

Yes. Thank you!