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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]  30 Sep 2010, 03:28
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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.
[Reveal] Spoiler: OA

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Re: Terminating Decimal [#permalink]  30 Sep 2010, 03:37
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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.

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

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]  17 Oct 2013, 07:39
Expert's post
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.

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]  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]  18 Nov 2014, 06:25
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Re: Any decimal that has only a finite number of nonzero digits   [#permalink] 18 Nov 2014, 06:25
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# Any decimal that has only a finite number of nonzero digits

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