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# If x is an integer, can the number (5/28)(3.02)(90%)(x) be

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If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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Updated on: 02 Nov 2012, 23:42
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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

Source: Jamboree

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Originally posted by Marcab on 02 Nov 2012, 23:40.
Last edited by Bunuel on 02 Nov 2012, 23:42, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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02 Nov 2012, 23:56
7
11
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.)

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
if-r-and-s-are-positive-integers-141000.html

BACK TO THE ORIGINAL QUESTION:

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.

Hope it's clear.
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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02 Nov 2012, 23:54
1
1
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.
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Re: finite number of non-zero digits  [#permalink]

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10 Jan 2013, 16:55
1
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.
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finite number of non-zero digits  [#permalink]

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10 Jan 2013, 11:10
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
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Joined: 09 May 2013
Posts: 43
Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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21 Jun 2013, 02:21
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.)

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
if-r-and-s-are-positive-integers-141000.html

BACK TO THE ORIGINAL QUESTION:

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.

Hope it's clear.

"Brilliant Bunuel"
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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11 Dec 2014, 03:20
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!!
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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11 Dec 2014, 06:41
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.
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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21 Apr 2015, 14:38
Marcab 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

Source: Jamboree

So when the question says CAN the number be a terminating decimal, it's actually asking is it ALWAYS a terminating decimal?

Posted from GMAT ToolKit
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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21 Apr 2015, 15:29
dqi2016 wrote:
Marcab 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

Source: Jamboree

So when the question says CAN the number be a terminating decimal, it's actually asking is it ALWAYS a terminating decimal?

Posted from GMAT ToolKit

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

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21 Apr 2015, 19:52
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?
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If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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22 Apr 2015, 00:37
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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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be  [#permalink]

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02 Jun 2016, 10:15
Marcab 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

Source: Jamboree

A number can be represented as infinite decimal if the denominator is in the form 2^n 5^m

Looking at the expression, (5/28)(3.02)(90%)(x), we are fine with all numbers except 28 as it is a multiple of 7. If any statement can tell us that either the expression gets rid of this 7 or it doesn't, we can conclude if the expression is terminating or not. (Basically look for statement that tells us if x is or not a multiple of 7)

(1) x is greater than 100. x may or may not be multiple of 7. Not sufficient
(2) x is divisible by 21. That means x is a factor of 7; and hence it will leave 4 as the decimal. we will get a terminating decimal. Sufficient

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

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05 Oct 2018, 10:01
Marcab 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

Source: Jamboree

$$\frac{5}{28} * \frac{302}{100} * \frac{90}{100} * x$$

for this to be a terminating decimal the denominator must be in the form of $$\frac{x}{2^a * 5^b}$$

Re-writing the question stem numbers to be $$\frac{5}{7*2^2} * \frac{302}{2^2 * 5^2} * \frac{90}{2^2 * 2^5} * x$$

from statement 1) if we x > 100 it might include a multiple of 7 and it might not. so insufficient.

2) x is divisible by 21, meaning it could be 21,42,63 etc.. it is a multiple of 21 and 21 = 7 * 3

This will cancel the 7 in the denominator and will lead to the terminating decimal form, sufficient.

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

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26 May 2019, 16:30
$$\frac{5}{28} *\frac{302}{100} * \frac{90}{100} * x$$

$$\frac{x*5*2*151*9*10}{2^{4}* 7 * 10^{4}}$$

$$\frac{x * 9 * 151}{2^{4} * 7 * 10^{2}}$$

Reworded Question: is x=7 or a multiple of 7?

(1) x is greater than 100
if x = 700 then YES
if x = 101 then NO
Insufficient.

(2) x is divisible by 21
If x is divisible by 21, then it's divisible by 7.
Sufficient.
Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be   [#permalink] 26 May 2019, 16:30
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