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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
02 Nov 2012, 22:56

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THEORY:

Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits?

First of all \(\frac{5}{28}*3.02*\frac{9}{10}*x=\frac{5*302*9*x}{28*100*10}=\frac{5*302*9*x}{7*(4*100*10)}=\frac{5*302*9*x}{7*(2^2*2^2*5^2*2*5)}\). Now, according to the theory above, in order this number to be termination decimal, 7 must be reduced by a factor of x (no other number in the numerator has 7 as a factor and all other numbers in the denominator have only 2's and 5's), so it'll be a terminating decimal if x is a multiple of 7.

(1) x is greater than 100. Not sufficient. (2) x is divisible by 21. Sufficient.

Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
02 Nov 2012, 22:54

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If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits? (1) x is greater than 100 (2) x is divisible by 21

Ok. WHat do we have at the denominator in fraction. 100 and 28. To be non recurring or finite decimal, Denominator should be made of multiple 2 or 5 or both. 100 is not a problem. 28 has 7 which is an issue.

B says X is divisible by 21. 7 goes in denominator. Hence Sufficient. _________________

Re: finite number of non-zero digits [#permalink]
10 Jan 2013, 15:55

1

This post received KUDOS

Expert's post

sambam wrote:

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits?

(1) x is greater than 100 (2) x is divisible by 21

Merging similar topics. Please refer to the solutions above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3: the name of a topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question. _________________

Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
21 Jun 2013, 01:21

Bunuel wrote:

THEORY:

Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits?

First of all \(\frac{5}{28}*3.02*\frac{9}{10}*x=\frac{5*302*9*x}{28*100*10}=\frac{5*302*9*x}{7*(4*100*10)}=\frac{5*302*9*x}{7*(2^2*2^2*5^2*2*5)}\). Now, according to the theory above, in order this number to be termination decimal, 7 must be reduced by a factor of x (no other number in the numerator has 7 as a factor and all other numbers in the denominator have only 2's and 5's), so it'll be a terminating decimal if x is a multiple of 7.

(1) x is greater than 100. Not sufficient. (2) x is divisible by 21. Sufficient.

Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
28 Aug 2014, 04:55

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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
11 Dec 2014, 02:20

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What happens if x has many 000s in the end so that there is no decimal left e.g. if x= 21000000 - there will no finite number of non-zero decimals. I think (b) is also not sufficient. Please help!!

Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
11 Dec 2014, 05:41

Expert's post

shilpabhagwat wrote:

What happens if x has many 000s in the end so that there is no decimal left e.g. if x= 21000000 - there will no finite number of non-zero decimals. I think (b) is also not sufficient. Please help!!

It's VERY easy to test that. Plug x = 21,000,000 and see what you get. _________________

Hello dqi. It can't be always or not always it's or can be represent or cannot be represent. If X doesn't contain 7 than this number can't be represent as terminate decimal. (always) If X contain 7 than this number can be represent as terminate decimal. (always) _________________

Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
21 Apr 2015, 18:52

But according to statement 1, if x>100, then if x is 280, that number would be a terminating decimal. But if x=101, then it's non terminating. So technically, abusing by statement 1, x "can" be a number to make it terminating, but isn't necessarily?

If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
21 Apr 2015, 23:37

dqi2016 wrote:

But according to statement 1, if x>100, then if x is 280, that number would be a terminating decimal. But if x=101, then it's non terminating. So technically, abusing by statement 1, x "can" be a number to make it terminating, but isn't necessarily?

I think now I correctly understand your first question about "always" Yes we should find statement (or both statements) that give us condition when X always make this number terminating. If statement give us variants - this is wrong statement.

In this task in second statement X always make this number terminating. And in first statement sometimes yes (x =280) - this number will be terminating and sometimes no (x = 101). This is general rule for any DS task. We should find condition (or combination) which always will give us needed result without any execeptions. _________________

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