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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? (1) k = 3 (2) j is an odd multiple of 3.
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Re: Terminating Decimal [#permalink]
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30 Sep 2010, 04:37
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 nonnegative 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 2s and/or 5s 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: 700question94641.html?hilit=terminating%20decimalisrs2isaterminatingdecimal91360.html?hilit=terminating%20decimalplexplain89566.html?hilit=terminating%20decimalwhichofthefollowingfractions88937.html?hilit=terminating%20decimalBACK 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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07 Oct 2010, 01:19
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 nonmultiple 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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07 Oct 2010, 01:36
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 nonmultiple 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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08 Oct 2010, 11:08
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 nonmultiple 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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17 Oct 2013, 05:49
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 nonnegative 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 2s and/or 5s 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: 700question94641.html?hilit=terminating%20decimalisrs2isaterminatingdecimal91360.html?hilit=terminating%20decimalplexplain89566.html?hilit=terminating%20decimalwhichofthefollowingfractions88937.html?hilit=terminating%20decimalBACK 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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17 Oct 2013, 08:39
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 nonnegative 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 2s and/or 5s 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: 700question94641.html?hilit=terminating%20decimalisrs2isaterminatingdecimal91360.html?hilit=terminating%20decimalplexplain89566.html?hilit=terminating%20decimalwhichofthefollowingfractions88937.html?hilit=terminating%20decimalBACK 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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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.



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