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Re: If r and s are positive integers [#permalink]
20 Oct 2012, 09:02

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.

Re: If r and s are positive integers [#permalink]
20 Oct 2012, 12:18

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. _________________

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Re: If r and s are positive integers [#permalink]
23 Oct 2012, 06:59

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asveaass wrote:

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 2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

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 (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 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.

Re: If r and s are positive integers [#permalink]
24 Dec 2013, 10:12

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?

Re: If r and s are positive integers [#permalink]
04 May 2014, 10:31

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?

Re: If r and s are positive integers [#permalink]
05 May 2014, 01:31

Expert's post

halloanupam wrote:

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?

0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit. _________________

Re: If r and s are positive integers [#permalink]
05 May 2014, 01:35

Expert's post

russ9 wrote:

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?

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. _________________

Re: If r and s are positive integers [#permalink]
26 Sep 2014, 22:28

Bunuel wrote:

asveaass wrote:

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 2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

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 (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.)

Why is it then 130/13 or 121/11 would give finite ... infact they properly divide... Can someone please help ?

Re: If r and s are positive integers [#permalink]
27 Sep 2014, 00:28

Expert's post

ani781 wrote:

Bunuel wrote:

asveaass wrote:

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 2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

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 (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.)

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. _________________