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

sorry Bunel: i will take care of that

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

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

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.

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.

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