C

(1) n is odd
if n is odd, then pick some numbers...
(n, (n-1)*(n+1), r)
(1, 0, 0)
(3, 8, 8)
INSUFFICIENT

(2)
(1, 0, 0)
(2, 3, 3)
INSUFFICIENT

Together, find some more examples...
(1, 0, 0)
(3, 8, 8)
(5, 24, 0)
(7, 48, 0)
(9, 56, 8)
So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0.
Together is SUFFICIENT

1.
3,5,7,9
n>=5
therefore 5,7,9

test 9
9^2-1=80
80/24=3R8
test 5
5^2-1=24

insufficient

2.
2,4,5,7,8
n>=5
therefore 5,7,8
test 8
8^2-1=63
63/24=2R15

insufficient due to testing 5 in 1

1 and 2)
5 and 7
test 7
7^2-1=48
48/24=2

test 5 had no remainder

C

Test numbers.

1: Try n=3 and n=23.

2: try n=2 and n=47.

Together r will be 0.

my thoughts -

N is not divisible by 2: Means n-1 and n+1 are even.
And - As they are consecutive numbers one of them is divisible by 2 and other by 4.
So: (n-1)* (N+1) is divisible by 8 - not sufficient to conclude that the product is divisible by 24

N is not divisible by 3: means as n-1,n,n+1 are consecutive , one of them: (n-1) or (n+1) is divisible by 3. Again, not sufficient for 24.

Together - (n-1)(n+1) is divisible by 2,4,3 i.e: 2*4*3 = 24 . Hence, r=0

does it make sense?

This is how I solved it
From statement 1) 2 is not a factor of n ( we can conclude that n is odd ) so by plugging in numbers like n=15 remainder=1 and for n=17 remainder =0 i.e not sufficient
From statement 2 it says that 3 is not a factor of n so we can have n as an even number or it can be prime number( not #0 again by plugging in values we can find that the remainders are different.i.e insufficient.

So by combining both we can conclude that the value for n will be a prime number which is not 2 and 3
for n=19 remainder =0 ; for n=13 remainder =0 ( just testing cases )

SO answer is C

Guys let me know if my solution is good ?

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?

Number plugging method:

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

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if n=1, then n^2-1=0 and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if n=3, then n^2-1=8 and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if n=1, then n^2-1=0, so remainder is 0 but if n=2, then n^2-1=3 and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3:
n=1 --> n^2-1=0 --> remainder 0;
n=5 --> n^2-1=24 --> remainder 0;
n=7 --> n^2-1=48 --> remainder 0;
n=11 --> n^2-1=120 --> remainder 0.
Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:

(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd --> n-1 and n+1 are consecutive even integers --> (n-1)(n+1) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either n-1 or n+1 must be divisible by 3 (as n-1, n, and n+1 are consecutive integers, so one of them must be divisible by 3, we are told that it's not n, hence either n-1 or n+1).

(1)+(2) From (1) (n-1)(n+1) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by 8*3=24, which means that remainder upon division (n-1)(n+1) by 24 will be 0. Sufficient.

Answer: C.

Hope it helps.
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