Stmt I

x+3 divisible by 3

x+3 / 3 = k (some integer)
x = 3k-3
= 3 (k-1)

k=1 x=0 yes
k=2 x = 3-> NO

k can be odd or even so may or may not be divisible by 6

INSUFF

Stmt II

x+3 -> odd

x=2 ->NO
x6-> YES

INSUFF

Together:

x = 3(k-1)
x +3-> odd => x = odd - 3(odd) we know odd-odd = even

For x to be even k-1 has to be divisible by 2 since 3 is not divisible by 2

x is always divisible by 6 since x = 3(k-1) is divisible by 3 and 2

SUFF

Choose C

0 and 12 -> violating the conditions in the statements. Can u please help me understand what u meant?

Regards,
CR

billyr wrote:Is the integer x divisible by 6?

(1) x+3 is divisible by 3

(2)x+3 is and odd number


the answer is c.(but i still think it is E since if we insert the number x with 0 and 12, it still not violate 1 and 2 statement, thus the answer is no because tehre are 2 answer)


Thanks

OA is correct,

from stmt1 : 3k+3 = 3(K+1) is divisible or not by 6 only when we know wot is k , hence not sufficient

from stmt2 , since 3 is odd x has to be even to make x+3 odd not alone sufficient

using both 1 & 2 x is the form of 2*3K hence divisible by 6 always

If X is devisible by 6 than it should be devisible by 2 and 3 both.

1) x+3 is divisible by 3
=> X+3 = 3K
=> X = 3(K-1)

so X is devible by 3 (eg. 6) but for the case of 9, it is not devisible by2.

2) x+3 is and odd number

x+3 = 2n+1 an odd no.
=> x = 2(n-2) => devisible by 2. but not necessarly by 3 (eg. 4)

combining both X|2 and X|3 => X|6. Ans C.

IMO C. But a slight different approach.

Question stem is asking " X divisible by 6? " or iwe can rephrase the question as "Is x divisible by 2 and 3? "

From stmt 1 - It is given x + 3 is divisbile by 3 ==> we can conclude that x is also divisible by 3. But we don't know whether x is divisible by 2 or not. Hence insufficient.

From stmt2 - x + 3 is odd ==> when the sum of x and an odd number is odd, then x should be even. In other words we can conclude that x is divisible by 2. Since every even number is divisible by 2. But we don't know if x is divisible by 3. Hence insufficient.

Now combining stmts 1 and 2. we know x is divisble by 2 and divisible by 3. Therefore, x is divisible by 6. Hence sufficient.