remainder DS question

The ans should be 'A' here.

Fro the q : y=x^2+8x+7=>y=(x+1)(x+7) (by factorizing)

From 1 : x could be 11, 17,23,29... and so on (series is with a diff. of 6)
since x=6k+5 (where, k is a multiplication factor and 6 is reminder)

So we can see the reminder is 0.
e.g. letx=11 =>y=(11+1)(11+7) which is completely divisible by 12 .

Try with rest of the no. you would get 0 as reminder.Suficient

From2: x is not a multiple of 3.so x could be 2,4,5,7 and so on.
reminder is different .So insufficient.

I think answer is A

(1)
When x is divided by 6 the remainder is 5 can be expressed in х=6q+5
insert this to main statement
y=x^2+8x+7 = (6q+5)^2+8(6q+5)+7 = 36q^2+108q+72 = 12(2q^2+9q+6)+0
remainder is 0
SUFFICIENT

(2)
x is not a multiple of 3
Here I pluged numbers
x=1 makes y=12, when 12 is divided by 12 remainder is 0
x=2 makes y=43, when 43 is divided by 12 remainder is 7
INSUFFICIENT

Answer is A