Data Sufficiency

From 1 n = 3,5,7,11
From 2 n=2,4,8,10

In each case r varies

Combined n = 5,7,11 etc
r unique =0 ..

Hence C

hi could u pls explain this one again? how did u find the possible values of n?
thanks in advance

from the stmts ..say 2 is not a factor of n ..so n cud be all odd numbers ..and likewise for the next

hope this helps

Thanks so much for ur immediate reply. Another question if u dont mind.

I got how u narrowed it down to n = any prime number greater than 3

but how did u figure out that the sq of a prime number - 1 is a multiple of 24 and so r = 0?

is there a rule or some thing,

thanks again

n is integer, r is the remainder when (n-1)(n+1) is divided by 24. r=?


24 = 2*2*2*3

so (n-1)(n+1) must have 2,2,2,3 i.e 8 & 3 as its factors to be evenly divisible by 24 or else it will give a remainder.

1) 2 is not a factor of n

i.e n is odd so both n-1 & n+1 will be even & since the two nos are consecutive multiples of 2, so if one is divisible by 2 one time, the other will be divisible by 2, 2 times

hence (n-1)(n+1) will have a factor of 8

still we cant say that 3 is one of the fators as n itself may be an odd multiple of 3

2) 3 is not a factor of n

i.e n gives a reamainder of 1,2 when divided by 3

in that case either n-1 or n+1 will have 3 as factor

but we still cant say that this product is divisible by 8

Combine

(n-1)(n+1) is divisible by 8

&
(n-1)(n+1) is divisible by 3

so (n-1)(n+1) is divisible by 24

r =0 SUFF

C

Samir

As always an amazing explaination. Thanks a lot