Divisibility by 3

[CyberSpy]

If x, y and k are integers, is xy divisible by 3?

(1) y = 2^16 - 1

(2) The sum of the digits of x equals 6k

SPOILER: OA: A

[adi_800]

How can OA be A?
Confused?

[CyberSpy]

statement 1 corrected..

I wonder why statement 2 is not sufficient.. if sum of the digits of x is of the form 6k then x should be divisible by 3 and so xy should also be divisible by 3 :S

[madboy]

It has to be D...

[adi_800]

Yup...D it is...

[code9063]

Is there a way to find out whehter condition 1 is sufficient without actually calculating the number?

[3Gs]

A is correct

what if k = 0...?

In that case x = 0, xy = 0 and xy/3 = 0/3 = 0

I think, its A and NOT D

[nverma]

Originally Posted by 3Gs

A is correct

what if k = 0...?

In that case x = 0, xy = 0 and xy/3 = 0/3 = 0

I think, its A and NOT D

3Gs has a given a perfect scenario..

Just adding to it, the sum of digits of X can only b zero if X is zero..

and zero*y = 0, and zero is a multiple of 3, or matter of the fact of every integer..

[EdYo]

IMO A

(1) y = 2^16 - 1 can be rewritten like this

(2^8 -1)*(2^8+1) = (2^4-1)*(2^4+1)*(2^8+1) = (2^2-1)*(2^2+1)(2^4+1)*(2^8+1) = 3*5*17*257

those are prime numbers and for sure x*y is divisible by 3
==> hence (1) suf --> eliminate C, E

(2) can't get anything from 6k because we don't know what is the value of k for sure. For example: if k = 3 --> sum of all digits of x is 63 which is 6+3 = 9 so divisible by 3. What is k = 1, 2, 0, 4 (would be faster to eliminate if you see 1 equation with 2 variable; 6k*y/3 =???) --> ins --> eliminate B, D

Hence the answer is A