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

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30 Sep 2010, 04:28

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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?

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.)

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 \(k\).

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!

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 _________________

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. _________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

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.)

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 \(k\).

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.)

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 \(k\).

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

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25 Oct 2013, 13:36

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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.

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

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18 Nov 2014, 07:25

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

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06 Mar 2016, 05:49

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