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Any decimal that has only a finite number of nonzero digits [#permalink]
20 Aug 2010, 12:18

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E

Difficulty:

25% (medium)

Question Stats:

70% (01:46) correct
30% (01:18) wrong based on 82 sessions

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?

Re: OG 11 Data Sufficiency 107 [#permalink]
21 Aug 2010, 10:40

tarn151 wrote:

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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Re: OG 11 Data Sufficiency 107 [#permalink]
21 Aug 2010, 10:55

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.. _________________

Re: OG 11 Data Sufficiency 107 [#permalink]
21 Aug 2010, 14:06

4

This post received KUDOS

Expert's post

tarn151 wrote:

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 onlyb (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]
01 Nov 2013, 00:25

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Re: Any decimal that has only a finite number of nonzero digits [#permalink]
18 May 2014, 09:53

1

This post received KUDOS

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

gmatclubot

Re: Any decimal that has only a finite number of nonzero digits
[#permalink]
18 May 2014, 09:53