for n=1, n squared -1 becomes 0 hence r is 0 which is also true for any odd n. Hence it works for 1

I think your question lies in, what happens when zero is divided by another number...what is the remainder?

When zero is divided by another number the remainder is zero; you get zero as an answer....nothing is left.

I have heard people explain it like this. You have a pizza and want to divide it up between 10 friends, how much pizza does each person get? They get nothing, because you had nothing...thus zero.

YTT wrote:Could anyone explain to me why the number 1 would work in the first situation? I understand why 3, 5, 7 or others work. But why 1 works, too? Thank you so much for this great help!!

n^2-1 = (n+1)(n-1)

if n is an odd num (except 1) , either n+1 or n-1 is a multiple of 4 and other a multiple of 2
for example if you take 17 => 17^2-1 = (17+1)(17-1) = 18*16 ; now 18 is a multiple of 2 and 16 is a multiple of 4 hence 17^2-1 is a multiple of 8. the same is true for any positive odd integer except 1

in the case of n^2 -1 = 0 and zero is a multiple of any number- zero is a multiple of 8

https://www.beatthegmat.com/is-zero-to-b ... t4268.html
https://www.beatthegmat.com/mba/2009/08/ ... torization (see the comments)

YTT wrote:Could anyone explain to me why the number 1 would work in the first situation? I understand why 3, 5, 7 or others work. But why 1 works, too? Thank you so much for this great help!!

Well,basically the two imp. concepts used in this ques. are
0 is an even number
0 is divisible by 8

Now, for n=odd we have 1 as an odd number.Also,1 is not divisible by 8 as it leaves the remainder as 1.
Now,zero divided by 8 gives 0 as the remainder.

So, it is clear that 1 can be taken as an answer choice.



Also,I wanted to ask in which option choice did u have the doubt:-For option C,when we take both the statements together ?? or do you have doubt in any other option choice ??

YTT wrote:Could anyone explain to me why the number 1 would work in the first situation? I understand why 3, 5, 7 or others work. But why 1 works, too? Thank you so much for this great help!!

well i would like to solve your problem this way
let odd no.be n=2k+1
so (2k+1)^2 -1 =4k^2 +4K +1 - 1 = 4k^2 +4K =4k(k + 1)
so it is clearly divisble by 4
Now either k or k+1 is even so it is divisible by 8 also

now coming for value n =1 it gives (n)^2 -1 = 0
so my friend YTT always remember 0 is divisible by any no. except itself
Hence A is suff.