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

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20 Aug 2010, 12:18

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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 numbers. If r and s are positive integers and the ratio is r/s is expressed as a decimal, is r/s a terminating decimal?

OK, so the question I have below has a very similar post on this forum (as well as other's I've checked) but it appears that my question is slightly different. With data point #2, every other forum post I've checked has s=4 for this question. In this case, the answer is B as any integer divided by 4 results in a terminating decimal. HOWEVER, in my OG11, I have s=4B (with no other reference to B anywhere in the question, as you see below). Please take a look, and let me know your thoughts.

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 numbers. If r and s are positive integers and the ratio is r/s is expressed as a decimal, is r/s a terminating decimal? 1. 90<r< 100 2. s = 4B

answer is still choice B in my version! I don't see how this is possible as once you introduce the other variable, s could really be an infinite number of positive integers

My understanding is as follows.

For (r/s) to be a terminating decimal, its denominator 's' should contain powers of 2 and/or 5 in it. Hence any denominator which can be expressed as (2^x * 5^ y), where x, y could be any positive integers.

In the two statements a) 90 < r < 100 --- This does not give us any clue about the denominator. Hence insufficient.

b) s= 4B. Now (r/s) becomes (r/(4s)) and the denominator is of the form (2^2 * 5 ^ 0 * s) and hence the division would result in a terminating decimal.

Hence statement is sufficient.

I don't see how this is possible as once you introduce the other variable, s could really be an infinite number of positive integers

Even after introducing a new variable B, s could be a infinite number of positive integers but still it would be of the form (2^ x * 5 ^y) and hence for all those infinite combination the fraction would be a terminating decimal.

Hope my explanation helps.
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Just to clarify your point s=4B, so it is NOT of the form 2^x 5^y, but will contain these factors for sure. If I understand right all it takes for a terminating decimal is to have the denominator contain 2 and/or 5 as factors..
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OK, so the question I have below has a very similar post on this forum (as well as other's I've checked) but it appears that my question is slightly different. With data point #2, every other forum post I've checked has s=4 for this question. In this case, the answer is B as any integer divided by 4 results in a terminating decimal. HOWEVER, in my OG11, I have s=4B (with no other reference to B anywhere in the question, as you see below). Please take a look, and let me know your thoughts.

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 numbers. If r and s are positive integers and the ratio is r/s is expressed as a decimal, is r/s a terminating decimal? 1. 90<r< 100 2. s = 4B

answer is still choice B in my version! I don't see how this is possible as once you introduce the other variable, s could really be an infinite number of positive integers

It must be a typo. The answer to the question if statement (2) says \(s=4B\) would be E, as we have no info about B.

As for the terminating decimals:

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.)

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

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01 Nov 2013, 00:25

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

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18 May 2014, 09:53

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IMO B, the decimal is terminating or not depends on the DENOMINATOR 1)90 < r < 100 but we need to know what is s. for example any number that when divided by 3 gives remainder will be non terminating.. ex-91/3, 94/3 etc

2)s=4 for any number which is divided by 4, remainder is 0,1,2or 3 for all of them you get terminating decimals

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

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14 Oct 2014, 15:01

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 numbers. If r and s are positive integers and the ratio is r/s is expressed as a decimal, is r/s a terminating decimal? 1. 90<r< 100 2. s = 4B

There is a mistake in the text. It says that any decimal number has only a finite number of nonzero digits, but this is not true. For this to be true, It must say: any decimal number has only a finite number of digits. Do you agree?

There is a mistake in the text. It says that any decimal number has only a finite number of nonzero digits, but this is not true. For this to be true, It must say: any decimal number has only a finite number of digits. Do you agree?

The reason it says "non zero digits" specifically is because theoretically every decimal has infinite trailing 0s at the right of the decimal after the last non zero digit.

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

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17 Dec 2015, 14:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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