sameerballani wrote:Q) If n is a positive integer and r is the remainder when (n+1)(n-1) is divided by 24, what is the value of r?
1)n is not divisible by 2
2)n is not divisible by 3

OA:C

Solution:
It can be easily verified that each statement alone is not sufficient to answer the question.
Now, combine both the statements together and check.
Since n is not divisible by either 2 or 3, it is of the form 6k+1 or 6k+5, k being any integer.
So, if n = 6k+1, (n+1)(n-1) = (6k+2)(6k) = 12k(3k+1). If k is even, 12k is divisible by 24 and hence 12k(3k+1) is divisible by 24.
If k is odd, 3k+1 is even and hence 12(3k+1)(k) is divisible by 24.
If n = 6k+5, (n+1)(n-1) = (6k+6)(6k+4) = 12(k+1)(3k+2).
If k is even, 3k+2 is even and hence 12(3k+2)(k+1)is divisible by 24.
If k is odd, (k+1) is even and hence 12(k+1)(3k+2) is divisible by 24.
Hence, we can say that if n is not divisible by both 2 and 3, (n+1)(n-1) is always divisible by 24 and the remainder is 0.

The correct answer is (C).