I get it as (B)

i get C too.

1) 2 not a factor if n=>n=odd=> (n-1) and (n+1) are consecutive evens
if they are 2 and 4 respectively, 2*4/24 leaves remainder r= 8
if they are 4 and 6, 4*6/24 leaves no remainder=>r=0
not sufficient


2) 3 not a factor of n=> n may be 2, 4, 5, 7, 8....
if n=2, (n-1)*(n+1)=1*3; 3/24 leaves remainder 3
if n=5, (n-1)*(n+1)=4*6; 24/24 leaves remainder 0
not sufficient

together,
n not a factor of 2 and 3=> n may be 5, 7, 11, 13...
if n=5, remainder=0 (from st-2 ab0ve)
if n=7, (n-1)*(n+1)=6*8; remainder=0
all other numbers give remainder 0

hence, C

Yup, (C).

aj5105 wrote:I get it as (B)

The question ask what is r when (n^2-1^2)/24
(n^2-1^2)/24 = (n^2 - 1)/24

1) 2 not a factor of n
if n = 9
(n^2-1^2)/24 = (81-1)/24 r = 9
if n = 11
(n^2-1^2)/24 = (121-1)/24 r = 1

2 different value of r, so not sufficient

2) same as 1) you will find different value of r
not sufficient

1) + 2)
If n is not factor of 3 and 2. e.g.
n = 11
n = 5
n = 7

r = 1 for (n^2-1^2)/24

thus, answer should be C

kic883 wrote:The question ask what is r when (n^2-1^2)/24
(n^2-1^2)/24 = (n^2 - 1)/24

1) 2 not a factor of n
if n = 9
(n^2-1^2)/24 = (81-1)/24 r = 9
if n = 11
(n^2-1^2)/24 = (121-1)/24 r = 1

2 different value of r, so not sufficient

2) same as 1) you will find different value of r
not sufficient

1) + 2)
If n is not factor of 3 and 2. e.g.
n = 11
n = 5
n = 7

r = 1 for (n^2-1^2)/24

thus, answer should be C

i think the remainder should be 0 for all the values mentioned above..