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Given that both x and y are positive integers, and that y =

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

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Given that both x and y are positive integers, and that y = 3^(x – 1) – x, is y divisible by 6?

(1) x is a multiple of 3

(2) x is a multiple of 4
[Reveal] Spoiler: OA

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Re: divisibility? [#permalink]

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New post 18 Jun 2011, 03:38
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AnkitK wrote:
Given that both x and y are positive integers and that y=3^(x-1)-x ,is y divisible by 6?
a.x is a multiple of 3
b.x is a multiple of 4


st 1: X can be 3 3^(3-1) - 3 = 9-3 = 6 yes divisible by 6
X can be 6 3^5 - 6 = 243-6 237 not divisble by 6

Hence not sufficient

St 2. X = 4 3^3 - 4= 23/6 = Not divisible by 6
X=8 3^7 - 8 = 2179/6 = not divisble by 6

hence sufficient
Its B

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Re: divisibility? [#permalink]

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New post 30 Jun 2011, 10:28
[/quote]

st 1: X can be 3 3^(3-1) - 3 = 9-3 = 6 yes divisible by 6
X can be 6 3^5 - 6 = 243-6 237 not divisble by 6

Hence not sufficient

St 2. X = 4 3^3 - 4= 23/6 = Not divisible by 6
X=8 3^7 - 8 = 2179/6 = not divisble by 6

hence sufficient
Its B[/quote]


Finding the factorial of 3^7, with all the additional simplification ..... will it be possible within 2 mins ?

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

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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.

Sufficient

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

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New post 15 Jun 2017, 03:27
For everyone picking numbers that let any expression get large, don't forget that 0 is a multiple of all 3 and 4 as well.

Instead of having to compute 3^5 - 6 to prove statement 1 to be insufficient, you can just compute 3^0 - 0 which gives you 1 which is not divisible by 6.

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

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New post 17 Nov 2017, 12:05
\(y = 3^{x – 1} – x\)

\(3^{x – 1}\) is odd, for \(y\) to be divisible by 6, \(x\) must be odd multiple of 3

question is reduced to x is odd multiple of 3?

Statement 1: \(x\) is multiple of 3
Not suff, x may be odd or even multiple of 3

Statement 2: \(x\) is multiple of 4
=> \(x\) is even
=> \(y = 3^{x – 1} – x\) => \(y = (odd - even) = odd\)
=> \(y\) is odd definitely not divisible by 6
=> Sufficient

Answer (B)

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

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New post 17 Nov 2017, 23:27
laxpro2001 wrote:
For everyone picking numbers that let any expression get large, don't forget that 0 is a multiple of all 3 and 4 as well.

Instead of having to compute 3^5 - 6 to prove statement 1 to be insufficient, you can just compute 3^0 - 0 which gives you 1 which is not divisible by 6.


Hi

The method of substituting 0 is cool, and I can see that how it can help in many questions, by making them easy.
However, here in this question, we Cannot substitute x=0 as its given both x and y are positive integers.

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Re: Given that both x and y are positive integers, and that y =   [#permalink] 17 Nov 2017, 23:27
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