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Any decimal that has only a finite number of nonzero digits

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Any decimal that has only a finite number of nonzero digits [#permalink]

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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) k = 3

(2) j is an odd multiple of 3.
[Reveal] Spoiler: OA

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Re: Terminating Decimal [#permalink]

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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:
700-question-94641.html?hilit=terminating%20decimal
is-r-s2-is-a-terminating-decimal-91360.html?hilit=terminating%20decimal
pl-explain-89566.html?hilit=terminating%20decimal
which-of-the-following-fractions-88937.html?hilit=terminating%20decimal

BACK TO THE ORIGINAL QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) \(k = 3\) --> now, if \(j=3p\) (j is a multiple of 3) then \(\frac{j}{k}=\frac{3p}{3}=p=integer=terminating \ decimal\) but if \(j\) is not a multiple of 3 then reduced fraction \(\frac{j}{k}=\frac{j}{3}\) won't be a terminating decimal, as denominator has primes other than 2 and/or 5. Not sufficient.

(2) \(j\) is an odd multiple of 3 --> \(j=3(2k+1)\), clearly insufficient as no info about the denominator \(k\).

(1)+(2) \(\frac{j}{k}=\frac{3(2k+1)}{3}=2k+1=integer=terminating \ decimal\). Sufficient.

Answer: C.
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Re: Terminating Decimal [#permalink]

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New post 07 Oct 2010, 00:19
But it says the ratio j/k is expressed as a decimal, so how come j=3p ?
I thought answer is A since J has to be a non-multiple of 3 since if it is a multiple of 3, j/k cannot be expressed as a decimal. COrrect me if i am wrong!
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Re: Terminating Decimal [#permalink]

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New post 07 Oct 2010, 00:36
psychomath wrote:
But it says the ratio j/k is expressed as a decimal, so how come j=3p ?
I thought answer is A since J has to be a non-multiple of 3 since if it is a multiple of 3, j/k cannot be expressed as a decimal. COrrect me if i am wrong!


Anser can't be A. Because just knowing that k=3, doesnt tell you much about the decimal. For instance if j and k do not have 3 as a common factor, it will not cancel out and you will not get a terminating decimal which you would if they do have 3 as a common factor.

Eg. j=1, k=3 : Decimal is 0.333333....
j=6, k=3 : Decimal is 2.0
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Re: Terminating Decimal [#permalink]

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New post 08 Oct 2010, 10:08
psychomath wrote:
But it says the ratio j/k is expressed as a decimal, so how come j=3p ?
I thought answer is A since J has to be a non-multiple of 3 since if it is a multiple of 3, j/k cannot be expressed as a decimal. COrrect me if i am wrong!


Can;t be true..
j can be a multiple of 3.
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Re: Terminating Decimal [#permalink]

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New post 17 Oct 2013, 04:49
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:
700-question-94641.html?hilit=terminating%20decimal
is-r-s2-is-a-terminating-decimal-91360.html?hilit=terminating%20decimal
pl-explain-89566.html?hilit=terminating%20decimal
which-of-the-following-fractions-88937.html?hilit=terminating%20decimal

BACK TO THE ORIGINAL QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) \(k = 3\) --> now, if \(j=3p\) (j is a multiple of 3) then \(\frac{j}{k}=\frac{3p}{3}=p=integer=terminating \ decimal\) but if \(j\) is not a multiple of 3 then reduced fraction \(\frac{j}{k}=\frac{j}{3}\) won't be a terminating decimal, as denominator has primes other than 2 and/or 5. Not sufficient.

(2) \(j\) is an odd multiple of 3 --> \(j=3(2k+1)\), clearly insufficient as no info about the denominator \(k\).

(1)+(2) \(\frac{j}{k}=\frac{3(2k+1)}{3}=2k+1=integer=terminating \ decimal\). Sufficient.

Answer: C.


So I guess for this type of questions we can never asume that k>j unless it says so in the question stem. Am I right?
Cheers
J :)
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Re: Terminating Decimal [#permalink]

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New post 17 Oct 2013, 07:39
jlgdr wrote:
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:
700-question-94641.html?hilit=terminating%20decimal
is-r-s2-is-a-terminating-decimal-91360.html?hilit=terminating%20decimal
pl-explain-89566.html?hilit=terminating%20decimal
which-of-the-following-fractions-88937.html?hilit=terminating%20decimal

BACK TO THE ORIGINAL QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) \(k = 3\) --> now, if \(j=3p\) (j is a multiple of 3) then \(\frac{j}{k}=\frac{3p}{3}=p=integer=terminating \ decimal\) but if \(j\) is not a multiple of 3 then reduced fraction \(\frac{j}{k}=\frac{j}{3}\) won't be a terminating decimal, as denominator has primes other than 2 and/or 5. Not sufficient.

(2) \(j\) is an odd multiple of 3 --> \(j=3(2k+1)\), clearly insufficient as no info about the denominator \(k\).

(1)+(2) \(\frac{j}{k}=\frac{3(2k+1)}{3}=2k+1=integer=terminating \ decimal\). Sufficient.

Answer: C.


So I guess for this type of questions we can never asume that k>j unless it says so in the question stem. Am I right?
Cheers
J :)


Yes, nothing in the stem indicates that k must be greater than j.
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Collection of Questions:
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: Any decimal that has only a finite number of nonzero digits [#permalink]

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New post 25 Oct 2013, 12:36
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This is a GMAT Hacks question of the day. The question reappeared on May 8, 2013. Here is the official explanation in case anyone was interested.

Answer: C Statement (1) is insufficient. If the denominator of the fraction is 3, the decimal would be terminating if the numerator is a multiple of 3. For instance, 6/3 = 2, a terminating decimal. However, if the numerator is not a multiple of 3, it will not be terminating, as in 7/3 = 2.33.

Statement (2) is also insufficient. The important factor in determining whether a fraction is equivalent to a terminating decimal is the denominator. If j = 9, the fraction could be 9/3 (terminating) or 9/7 (not terminating).

Taken together, the statements are sufficient. j/k is equal to (3(integer))/3 = integer. An integer is, as defined in the question itself, a terminating decimal. Choice (C) is correct.


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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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New post 16 Jan 2018, 16:55
rxs0005 wrote:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) k = 3

(2) j is an odd multiple of 3.


We need to determine whether j/k is a terminating decimal given that j and k are positive integers. One thing we should keep in mind is that a fraction (in lowest terms and with a denominator greater than 1) can be expressed as a terminating decimal if and only if the denominator comprises prime factors of only 2 and/or 5. For example, 3/10 and 3/15 = 1/5 are terminating decimals, whereas 3/7 and 3/9 = 1/3 are not. On the other hand, if the denominator is 1, the fraction is always a terminating decimal as long as the numerator is an integer.

Statement One Alone:

k = 3

Depending on the value of j, j/k may or may not be a terminating decimal. For example, if j = 1, then j/k = 1/3 is not a terminating decimal. On the other hand, if j = 3, then j/k = 3/3 = 1/1 = 1 is a terminating decimal. Statement one is not sufficient to answer the question.

Statement Two Alone:

j is an odd multiple of 3.

Depending on the value of k, j/k may or may not be a terminating decimal. For example, if j = 3 and k = 7, then j/k = 3/7 is not a terminating decimal. On the other hand, if j = 3, and k = 3, then j/k = 3/3 = 1/1 = 1 is a terminating decimal. Statement two is not sufficient to answer the question.

Statements One and Two Together:

Since j is an odd multiple of 3, and k = 3, j/k is always an odd integer. Thus, j/k is a terminating decimal.

Answer: C
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Re: Any decimal that has only a finite number of nonzero digits   [#permalink] 16 Jan 2018, 16:55
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