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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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18 Dec 2012, 05:27
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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4
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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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18 Dec 2012, 05:32
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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 $$b$$ (the 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 QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100. Nothing about the denominator. Not sufficient.

(2) s = 4. According to the above, any fraction r/4=r/2^2 when expressed as a decimal will be a terminating decimal. Sufficient.

Questions testing this concept:
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

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

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18 Dec 2012, 05:33
A fraction r/s will only be a terminating decimal ONLY if it is of the form $$Numerator/ 2^m 5^n$$, where n and m are non-negative.
Statement 1 gives the range of numerators, of which we are not concerned at all. Insufficient
Statement 2 gives the value of denominator which is of the form $$2^2$$. Hence the fraction has to be a terminating decimal.
+1B
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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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18 Dec 2012, 05:33
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Carcass rightly said, you are a machine Bunuel.
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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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29 Jul 2013, 18:43
Question, I understand that a terminating decimal has to be of the form $$2^x5^x$$ but four is only in the form of $$2^n$$ to be a terminating decimal it can meet either of the requirements?

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

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29 Jul 2013, 18:51
hfbamafan wrote:
Question, I understand that a terminating decimal has to be of the form $$2^x5^x$$ but four is only in the form of $$2^n$$ to be a terminating decimal it can meet either of the requirements?

Thanks,
Hunter

for a fraction to be terminating two condition must satisfy:
1) numerator is an INTEGER.
2) denominator should be of form $$2^x 5^y$$ $$(x,y$$==>integers which also includes 0)

now in this question
denominator is $$2^2 5^0$$
hence it satisfies.

hope it helps
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23 Aug 2014, 10:42
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08 Sep 2015, 04:40
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Re: Any decimal that has only a finite number of nonzero digits   [#permalink] 08 Sep 2015, 04:40
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# Any decimal that has only a finite number of nonzero digits

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