test with different odd numbers. in each of the cases, there will be no remainder. thus (1) is sufficient.
if you test with 2,4 or 6, which are not divisible by 8, you will get different remainders. thus (2) is insufficient.
Remainder - Data Sufficiency
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cramya
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Algebric/conceptual approach:
What is the remainder when n^2-1 / 8
or
What is the remainder when (n-1) (n+1) / 8
Stmt I
n is odd
When n is odd both n-1 and n+1 are 2 consecutive even integers. The product of any 2 consecutive even integers is always disivible by 8. The remainder is 0
2*4
4*6 ... so on
(Also note one of the 2 consecutive even integer is always divisible by 4 in a product of 2 consecutive even integers)
SUFF
Stmt II
n is not divisible by 8
n=2
(n-1) (n+1) / 8 = (1*3)/8 remainder= 3
n=5
(n-1) (n+1) / 8 = (4*6)/8 remainder= 0
INSUFF(many different values possible two of which are listed above)
Choose A
Hope this helps!
Regards,
CR
What is the remainder when n^2-1 / 8
or
What is the remainder when (n-1) (n+1) / 8
Stmt I
n is odd
When n is odd both n-1 and n+1 are 2 consecutive even integers. The product of any 2 consecutive even integers is always disivible by 8. The remainder is 0
2*4
4*6 ... so on
(Also note one of the 2 consecutive even integer is always divisible by 4 in a product of 2 consecutive even integers)
SUFF
Stmt II
n is not divisible by 8
n=2
(n-1) (n+1) / 8 = (1*3)/8 remainder= 3
n=5
(n-1) (n+1) / 8 = (4*6)/8 remainder= 0
INSUFF(many different values possible two of which are listed above)
Choose A
Hope this helps!
Regards,
CR
