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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] New post 02 Nov 2012, 22:40
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 02 Nov 2012, 22:42, 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] New post 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.)

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.
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Re: finite number of non-zero digits [#permalink] New post 10 Jan 2013, 15:55
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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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink] New post 02 Nov 2012, 22:54
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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finite number of non-zero digits [#permalink] New post 10 Jan 2013, 10:10
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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Re: If x is an integer, can the number (5/28)(3.02)(90%)(x) be [#permalink] New post 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.)

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