Hi,
N =2k+1.
So, N^2 = (2k+1)^2 = 4k^2+4k+1 = 4k(k+1) + 1
k,(k+1) are 2 consecutive integers. So, one of them is even and other is odd. So, product is even
So, 4k(k+1) = 4*even = multiple of8
So, N^2 when divided by 8 leaves remainder 1

Hence, A

Frankenstein wrote:Hi,
N =2k+1.
So, N^2 = (2k+1)^2 = 4k^2+4k+1 = 4k(k+1) + 1
k,(k+1) are 2 consecutive integers. So, one of them is even and other is odd. So, product is even
So, 4k(k+1) = 4*even = multiple of8
So, N^2 when divided by 8 leaves remainder 1

Hence, A

Good, although slightly overkill. A few plug ins can show the same pattern. N can only be odd integers:

if N=1, then N^2 = 1: leaves a remainder of 1 from the nearest multiple of 8 (which is zero)
if N=3, then N^2 = 9: leaves a remainder of 1 from the nearest multiple of 8 (which is 9)
if N=5, then N^2 = 25: leaves a remainder of 1 from the nearest multiple of 8 (which is 24)
if N=7, then N^2 = 49: leaves a remainder of 1 from the nearest multiple of 8 (which is 48)

Now it's true that four plug ins do not prove a rule indefinitely - it is possible that somewhere down the line, there is an odd example of N that does not yield a remainder of 1 when divided by 8. But usually we'll see some sort of break from the pattern before reaching 4 consecutive successes. For the purposes of the GMAT, these plug ins are enough to choose A with a high degree of certainty.

n=2m+1
n^2 = 4m^2 + 4m + 1
n=1;remainder=1
n=3;remainder=1
n=5;remainder=1
IMO A