If r and s are positive integers, can the fraction r/s be expressed as

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:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

BACK TO THE ORIGINAL QUESTION:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

(1) s is a factor of 100. Factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100. All these numbers are of the form \(2^n5^m\) (for example 1=2^0*5^0, 2=2^1*5^0, ...), therefore no matter what is the value of r, r/s will always will be terminating decimal. Sufficient.

(2) r is a factor of 100. We need to know about the denominator. Not sufficient.

Answer: A.

Hope it's clear.
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Finite decimals are decimals which end. eg 0.5,0.25, etc. Non finite are numbers like 1/3,1/6 etc ie 0.33333333333333333333333333333333333333333333333333333.......... or 0.66666666666666666666666666666666666666666666666666666..........

1)S is a factor of 100. So S cannot have more than two 2s and two 5s. Any number divisible be 2 or 5 gives a finite decimal. Since R and S are both positive integers, there can be no 0s in the decimal places either. So Sufficient

2)R is a factor of 100. Cant say anything form this. R can be 1. 1/10 is finite. 1/3 is not. Insufficient.

Answer is hence A.

IMO it is A, if the denominator is a factor of 100 then it could be 1; 2; 5; 10; 20.. if you divide all the positive integer by these number you will have a finite decimal result.

you rock bunuel, your explanations are awesome.. thanks a lot for this one!
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But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?

Hi Bunuel,

Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?

If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?

0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit.
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Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
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Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
Can someone please help ?

The rule above is for reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term). When you reduce 130/13 to the lowest term you get 10 and when you reduce 121/11 you get 11: 10/(2^0*5^0) and 11/(2^0*5^0) respectively.

Check the links in my post above to practice more on this type of questions.
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Hi Bunuel,

Just looking for some clarification, in the highlighted text above, please correct me if I am wrong, by including the word "and" did you mean that if the denominator contains only 2's or only 5's then too the fraction is a terminating decimal? In other words, the denominator doesn't need to contain, both - 2's & 5's!

Thank you!
dmmk

dmmk

Yes, examples such as \(\frac{17}{2^7}\), \(\frac{31}{5^{18}}\), and \(\frac{21}{[(2^5)(5^7)]}\) are all terminating decimals.

Dabral

Bunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?

Ergenekon

For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, \(\frac{21}{[(2^4)(5^3)(11^2)]}\), \(\frac{11}{[(5^6)(7^3)]}\), and \(\frac{22}{(7^4)}\) are all non-terminating decimals.

Dabral

Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?

An integer IS a decimal with a finite number of nonzero digits. For example, integer 51 has 2 (finite) number of digits.
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Hi Bunnel,

A couple of questions:

1) In a DS question how would you decide whether a fraction is reduced?

2) If the denominator has a prime factor other than 2 and/or 5 what should be the approach.With what i percieve from your previous explanations, in such a case we would have to check whether the fraction is reduced.If the perception is correct how to decide that.

3) Is this a universal rule ?

4) How would such rules click in the exam?


Regards

Hello kirtivardhan

1) In a DS question how would you decide whether a fraction is reduced?
In any question fraction can be reduced if nominator and denominator have a common factors.
\(\frac{6}{10}\) is not reduced fraction because \(6\) and \(10\) have common factor \(2\). We should divide both parts on \(2\) and receive \(\frac{3}{5}\)

2) If the denominator has a prime factor other than 2 and/or 5 what should be the approach.With what i percieve from your previous explanations, in such a case we would have to check whether the fraction is reduced.If the perception is correct how to decide that.
Firstly you should reduced fraction as described in previous clause. Secondly you should check if denomitaor has primes others than \(2\) and \(5\).

3) Is this a universal rule ?
Yes

4) How would such rules click in the exam?
I haven't take real exam but meet such questions like 1 times per two exams (CATs).
Usually it's quite simple questions so you should know this principle, because if you lose such question, you'll lose overall complexity of next questions and this will be have negative impact on your final score.
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Hi

In the question p/q, how can we be sure whether this can be reduced or it is already reduced

Also, if the first option would have been a multiple of 100 then the option would have been insufficient right? Since we could have any other prime factor