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Any decimal that has only a finite number of nonzero digits [#permalink]
18 Dec 2012, 04:27

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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?

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
18 Dec 2012, 04:32

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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 onlyb (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.

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
18 Dec 2012, 04:33

Expert's post

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 _________________

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
29 Jul 2013, 17:43

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?

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
29 Jul 2013, 17:51

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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Re: Any decimal that has only a finite number of nonzero digits [#permalink]
23 Aug 2014, 09:42

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