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Given that both x and y are positive integers, and that y =
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17 Jun 2011, 07:12
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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
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Re: Given that both x and y are positive integers and that
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14 Feb 2014, 19:36
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: divisibility?
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18 Jun 2011, 03:38
AnkitK wrote: Given that both x and y are positive integers and that y=3^(x1)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^(31)  3 = 93 = 6 yes divisible by 6 X can be 6 3^5  6 = 2436 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?
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30 Jun 2011, 10:28
[/quote]
st 1: X can be 3 3^(31)  3 = 93 = 6 yes divisible by 6 X can be 6 3^5  6 = 2436 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 ?
Regards, Mustu



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Re: Given that both x and y are positive integers, and that y =
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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 =
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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 =
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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.




Re: Given that both x and y are positive integers, and that y = &nbs
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17 Nov 2017, 23:27






