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Re: finite number of non-zero digits [#permalink]
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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]
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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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
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.

Answer: B.

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]
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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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]
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]
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]
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]
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]
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]
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.

Answer choice B
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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink]
\(\frac{5}{28} *\frac{302}{100} * \frac{90}{100} * x\)

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

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

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.
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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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