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Re: If x, y and k are integers, is xy divisible by 3? [#permalink]
20 Sep 2009, 18:55
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
I believe the correct answer is D.
Basically, in order for xy to be divisible by 3, x or y HAVE TO be divisible by 3 on their own, since 3 is a prime number.
Statement 1: y = 2^(16) - 1: The key here is knowing that 2^(odd integer) = 3n + 2, and 2^(even integer) = 3n + 1 (where n is a non-negative integer). Since 16 is an even integer, 2^16 will be equal to 3n + 1. Therefore 2^16 - 1 = 3n + 1 - 1 = 3n. Therefore y is divisible by 3, and Statement 1 is sufficient.
The reason that this rule works for 2^n when considering divisibility by 3 is as follows (sorry if this is confusing): if you take a number in the form of 3n + 1, and multiply it by 2, you get 6n + 2, which can be rewritten in the form 2*3n + 2 = 3m + 2. Multiplied again by 2, you get 6m + 4, which can be rewritten in the form 6(m+1) + 1 = 3p + 1. The number 2 (being 2^1) is in the form 3n + 2. Subsequently all numbers in the form 2^(odd number) can be written as 3n + 2, and all numbers in the form 2^(even number) can be written as 3n + 1. I remember learning this somewhere years ago, there might be a better explanation out there on the internet.
Statement 2: The sum of the digits of x equals 6^k : For x to be divisible by 3, the sum of its digits must also be divisible by 3. It is given that the sum of the digits of x equals 6^k.
\(6^k = 3^k*2^k\) and therefore x is definitely divisible by 3. Therefore Statement 2 is also sufficient.
Since both Statement 1 and 2 are both sufficient by themselves, the correct answer is D.
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