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Re: Given that both x and y are positive integers and that [#permalink]

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14 Feb 2014, 19:36

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A much easier approach:

In order for the expression to be divisible by 6 it must satisfy that it is divisible by 2 and 3.

Another way to view divisibility by 2 is Even/Odd, so the expression must be even to be divisible by 6.

S1. Since 3 to any power will always be odd, the other part of the expression (+X) must be odd for the expression to be even, and possibly divisible by 6. Since X is a multiple of 3 is the constraint, this is satisfied by both even and odd numbers, making the expression even or odd, depending on the value. It will be divisible by 6 when X is odd, given that (3^?) would be a multiple of 3 and so would be (+X) and it will be even.

Not sufficient.

S2. From the conclusion above, and since now we are told that (+X) is a multiple of 4, we now know that (+X) will ALWAYS be even, making the expression never divisible by 2 and by extension, never divisible by 6.

Re: Given that both x and y are positive integers, and that y = [#permalink]

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07 Oct 2015, 08:56

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Re: Given that both x and y are positive integers, and that y = [#permalink]

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19 Oct 2016, 11:21

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