A First thought it might be E. But then only noticed its n^2 minus 1 !! he he

Ian Stewart wrote:

zagcollins wrote:If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?

1)n is odd
2)n is not divisible by 8

You might notice that this is a difference of squares:

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

If n is odd, then n-1 and n+1 are consecutive even integers. If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8. Thus, if we know n is odd, we can be certain that n^2 -1 will be divisible by 8, and r will be zero. 1) is sufficient. 2) is not; n might be even, or might be odd. A.

The only exception to this is 1. But that won't trouble the answer in any way, because in that case, the product will be 0 - (1-1)*(1+1) = 0 thus choice A will hold true.

1. If n is odd integer, n = 2k + 1
We have : n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k = 4k(k + 1) is multiple of 8. So r = 0 : Sufficient ---> AD

2. If n = 1 : r = 0
If n = 2 : r = 3
Hence, (2) is insufficient ---> A is correct anwser.

For the expression (n^2)-1/8

If n is odd, remainder will always be 0

for e.g. 3^2 -1=9-1=8/8. Rem =0
5^2-1=24/8=3. Rem =0

13^2=169-1=168/8/. Rem=0

Thus option 1 is sufficient to answer the question.

Coming to Option 2,
2. n is not divisible by 8

This is not sufficient as n can be any number.

for e.g (9^2) -1=80/8. Rem =0
(10^2) -1=99/8. Rem =3
Thus option 2 cannot provide us the correct answer for the question.

Hence, option 1 is the answer.

Dear mana6,

while accessing the first statement you are taking even values for n, whereas the statement states that n is odd.

Answer is A.

1. If n is odd number, it can be 3, 5, 7, 9, 11, ...
so n^2-1 will be 8, 24, 48, 80, 120,...all are multiple of 8. Sufficient

2. For n = 2, 3...r will be 3, 0 Insufficient

Only statement number 1 is sufficient.

Answer: A

zagcollins wrote:If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?

1)n is odd
2)n is not divisible by 8

If n is odd it can be expressed as 2k+1 for any integer k.
So, n^2-1 = (2k+1)^2-1 =4k^2+4k=4k(k+1)=8m (where m is another integer)=> r=0 for all odd n.

A for answer.

Ans is A

just 10 second to answer : A.
n^2 always even integer
so n^2- 1 is an odd.
odd/8(even integer)= odd No.

The majority is right
The answer is A

Clearly `a` is the answer

Why aren't the answer choices listed? How am I supposed to choose an answer if infant see the choices? Please help!!!!

a. if n is odd then n must be 1,3, 5,7,9 then the last digit of n^2 and (n^2-1) must be 1,9,5,9 and 0,8,4,8.
Let (n^2-1) / 8 means that (n^2-1): 2:2:2.
we have the number with last digit 8 is always divisible by 2 ( for last digit only 8:2=4:2=2:2=1) ; the number with last digit 0 will have remainder 1 after dividing by 2:2:2, the number with last digit 4 will have remainder 1 after dividing by 2:2:2.
then the remander will be always 1, A is sufficient
b.n is not divisible by 8 then n can be odd or even => B is insufficient.

the answer is A