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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.

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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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First recognize that the tenths digit of x will equal ZERO, if x is LESS THAN 0.1 For example, if x = 0.04, the tenths digit is 0 So, the tenths digit of x will be NONZERO if x > 0.1 In other words, the tenths digit of x will be NONZERO if x > 1/10 Since we're already told that x < 1, we can REPHRASE the target question... REPHRASED target question:Is 1/10 < x < 1?

Statement 1: 16x is an integer. Since we're told that 0 < x (i.e., x is positive), we know that 16x is positive So, let's say that 16x = k, where k is some positive integer Solve for x to get: x = k/16 (where k is some positive integer) There are several values of k that satisfy statement 1. Here are two: Case a: k = 8, in which case x = 8/16 = 1/2, which means it IS the case that 1/10 < x < 1 Case b: k = 1, in which case x = 1/16, which means it is NOT the case that 1/10 < x < 1 Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 8x is an integer. Let's say that 8x = j, where j is some positive integer Solve for x to get: x = j/8 (where j is some positive integer) Since j is a positive integer, then j = 1 or 2 or 3 or .... In all of these cases, j/8 will be GREATER than 1/10 In other words, x must be GREATER than 1/10 Since we're also told that x < 1, we can be certain that 1/10 < x < 1 Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

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.

The question is an application of the concept of terminating decimals. Recall that when the denominator of a fraction has only 2s and/or 5s as factors, it is a terminating decimal.

x could be 0.345 or 0.02 etc. To create integers, you need to move the decimal to the right. This is done by making 10s. 16 and 8 in the statements should remind you of powers of 2. This should make you think of powers of 5 to make 10s.

(1) 16x is an integer.

Your have four 2s here. When you multiply them by four 5s (i.e. 625), you can make four 10s.

So if x = 0.625, then 16x will be an integer But if x = 0.0625, then also 16x will be an integer Not sufficient.

(2) 8x is an integer.

Here you have three 2s. When you multiply them by three 5s (i.e. 125), you can make three 10s.

So if x = 0.125, then 8x will be an integer. What about when x = 0.0125? No, 8x will not be an integer. We can make three 10s but 125 itself has 3 digits. You cannot have an additional 0 digit immediately after the decimal point. So we know that tenths digit of x cannot be 0.

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

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29 Sep 2016, 00:57

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.

i think it is easier.

to check if (1) is sufficient just think about the shortest number which make 16 an integer. it will be surely \(x=\)\(\frac{1}{16}\)=0.0...something so (1) is not sufficient

with the same logic the shortest number which make 8 an integer is \(x=\frac{1}{8}\)=0.1...something therfore statement (2) is sufficient.

answer B
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