If x, y and k are integers, is xy divisible by 3? (1) y =

Question

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 6^k

Bunuel · Accepted Answer

shrive555 please check the questions when posting. The correct question is as follows:

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

Now,  xy  will be divisible by 3 if either unknown is divisible by 3.

(1) y = 2^16 - 1 -->  y =2^{16}-1=(2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=15(2^4+1)(2^8+1) , so  y  is a multiple of 3. Sufficient.

(2) The sum of the digits of x equals 6^k --> if  k\geq{1}  then  x  is divisible by 3 (as the sum of the digits of  x  will be multiple of 3) and the answer to the question will be YES but if  k=0  then  x  could be 1 and we won't be sure whether  x  is a multiple of 3. Not sufficient.

Answer: A.

ankitranjan · Answer

(1) says y=215 Now xy=215*x So it depends on x whether is divisible or not ...> Insufficient
(2) says x is divisible by 3( condition of divisibility by 3 is the sum of the digits shud be divided by 3 and here it is 6k.)

Hence answer shud be B.

Consider KUDOS if its helpful.

shrive555 · Answer

sorry Bunel: i will take care of that

Eden · Answer

I need help understanding the red part.
I thought that if k=0, then the sum of digits of x will be zero (since the sum is a function of k), which implies that x is 0.
Maybe I am not reading the prompt correctly.
Thanks

subhashghosh · Answer

@Eden, any real number raised to 0 will be = 1, and 1 is not a multiple of 3.

vivesomnium · Answer

1. We can observe a pattern. 
2^1 - 1 = 1 Not divisible
2^2 - 1 =3 Divisible
2^3 - 1 = 7 Not Divisible
2^4 - 1 =15 Divisible
2^6 - 1 = 63

2 rasied to even power minus one is always divisible by 3. Hence y= 2^16-1 is divisible. If y is divisible by 3, so is xy. Hence sufficient

2. If sum of digits of x is equal to 6^k, where k is an integer, k can be 0 or a positive integer. (cant be negative as that will lead to a fractional value for 6^k which cant be possible as the sum of the digits) now if k is a positive integer- 6^k is always going to be divisible by 3, and since sum of digits is divisible by 3, so will be x, and hence so will be xy. But what if k is 0? Then 6^0 becomes 1, and all info (2) gives us is that sum of digits of x is divisible by 1 - so what ? all numbers are divisible by 1. this cant give enough information to determine whether xy is divisible by 3 or not. hence insufficient.

Spidy001 · Answer

Eden,

when k>=1, statement 2 is true and divisible by 3.
but when k=0 ,the sum of the digits equals to 6^0 = 1 ( not zero) , which is not divisible by 3. Thats why option 2 is not sufficient as we dont know whether k >=1 or k =0 or k<0.

bumpbot · Answer

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If x, y and k are integers, is xy divisible by 3? (1) y =

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If x, y and k are integers, is xy divisible by 3? (1) y =

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