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

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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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you rock bunuel, your explanations are awesome.. thanks a lot for this one!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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 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.)

Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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 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.)

Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...

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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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
Bunuel wrote:
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.

Bunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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.

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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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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?
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ptanwar1 wrote:
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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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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
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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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kirtivardhan wrote:
Hi
In the question p/q, how can we be sure whether this can be reduced or it is already reduced

If we have only this information, we can't say nothing about state of this fraction.
And if we have information only about $$r$$ or about $$q$$ it's the same: we can't say nothing about reducing of this fraction.

because in this case we need to know if denominator contain any other factors except $$2$$ and $$5$$.
and we already have information that denominator is a factor of $$100$$, so we know that denominator doesn't conatain any factors except $$2$$ and $$5$$ and we don't need to reduce fraction.

kirtivardhan wrote:
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

Absolutely right.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
''as a decimal with only a finite number of nonzero digits'' what does it exactly mean? Are 3.05, 9.002 etc under this category or sth else?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
oishik wrote:
''as a decimal with only a finite number of nonzero digits'' what does it exactly mean? Are 3.05, 9.002 etc under this category or sth else?

Yes, because there is a finite number, respectively 2 and 3, of nonzero digits after the decimal point. For example, 1/3 = 0.33333... is not a decimal with only a finite number of nonzero digits, because 3's there goes infinitely.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
What if r=25 and s=25

Then 25/25=1

It has no decimal point in this case.

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