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Given: \(x\) is a fraction in the range \(0<x<1\) --> \(x=0,abcd...\) Question: is \(a=0\)?

(1) 16x is an integer --> \(x=\frac{integer}{16}\), where \(0<integer<16\) (as \(0<x<1\)). Now, the least value of \(x\) is \(x=\frac{1}{16}=0.0625\) and the tenth digit is zero BUT the highest value of \(x\) is \(x=\frac{15}{16}=0.9375\) and the tenth digit is nonzero (9). Not sufficient.

(2) 8x is an integer --> \(x=\frac{integer}{8}\), where \(0<integer<8\) (as \(0<x<1\)). Now, the least value of \(x\) is \(x=\frac{1}{8}={0.125}\) and the tenth digit is already nonzero thus all other values of \(x\) will be more than 0.125 and therefore will have nonzero tenth digit. Sufficient.

Re: In the decimal representation of x, where 0 < x < 1, is the [#permalink]

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20 Sep 2012, 05:28

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An alternate & probably easy solution for this problem

1) x can be 1/2,1/4,1/8,1/16 & all will give integer when multiplied by 16. X can be = .5, .25, .125 or .0625 –Tenth digit can be zero but its not sure - Not Sufficient 2) x can be 1/2,1/4,1/8, & all will give integer when multiplied by 16. X can be = .5, .25, .125 –Tenth digit is always non zero – Sufficient Answer B

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Re: In the decimal representation of x, where 0 < x < 1, is the [#permalink]

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24 Jun 2013, 14:09

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Statement 1: For tenth digit to to be zero the fraction multiplied by 16 must be less than 1/10. For 16x to be an integer, x could be 1/16 or 1/8. Not Sufficient

Statement 2: Same logic as above except for 8 to be an integer it can only be multiplied by 1/8 (which is more than 1/10) Sufficient.

Re: In the decimal representation of x, where 0 < x < 1, is the [#permalink]

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25 Jun 2013, 10:11

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udaymathapati wrote:

In the decimal representation of x, where 0 < x < 1, is the tenths digit of x nonzero?

(1) 16x is an integer. (2) 8x is an integer.

From the given statement we can conclude that x is terminating decimal. Let x = .abc where we need to find whether a equal to 0 or not

From St 1 : 16x is an Integer ------->x can be 0.125 or 0.0625 as when multiplied by 16 both the values of x give us Integer values 2 and 1 respectively

From St 2: 8x is integer. Knowing that 8X is an integer we can safely conclude that "a" not equal to 0. The least possible value of x will be 0.125 and other possible values of x can be multiple of 0.125 only i.e 0.250, 0.375, 0.500 and hence "a" will never be zero.

Ans is B _________________

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Re: In the decimal representation of x, where 0 < x < 1, is the [#permalink]

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15 Sep 2014, 01:04

udaymathapati wrote:

In the decimal representation of x, where 0 < x < 1, is the tenths digit of x nonzero?

(1) 16x is an integer. (2) 8x is an integer.

Basically this question is asking whether x lies in 0 < x < 0.099999 or not ?

1) 16x is an integer.

multiply 16 x 0.099 = 1.44 << it means we can have some value less then 0.099 at which product with 16 yields 1. further 16x0.5=8 is also integer. Therefore, at some value 0.0xy and 0.5 I am getting integer value and thus, 1 is not sufficient.

2) 8x is an integer. multiply 8 x 0.099 = 0.72; we can see that 0.72 is less than nearest integer 1 and thus we will have to multiply 8 with some number greater than 0.099 and that number will definitely not have 0 in tenths place. e.g x = 1/8 = 0.125 Therefore Statement 2 is sufficient.

Ans B. _________________

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18 Apr 2016, 05:58

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