OG 11 - Properties of numbers

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OG 11 - Properties of numbers

by albertrahul » Mon Jul 07, 2008 5:26 am
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 decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

(1) 90 < r < 100
(2) s = 4

Please explain your solution, I'm posting this question here because
(a) OG explanation did not make sense to me.
(b) I want to know how to approach these type of problems.
Source: — Data Sufficiency |

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by lion147 » Mon Jul 07, 2008 7:28 am
I think the answer is B.

1) If r is 91 and s could be any number, then 91/3 = 30.333... - a non-terminating decimal, or r could be 93 and s 3, then 93/3 = 31.0 - a terminating decimal.

2) When you divide any integer by 4, the only possible trailing numbers including the decimal point are .0, .25, .5 and .75, all of which are terminating decimals.

e.g.
1/4 = 0.25
2/4 = 0.5
3/4 = 0.75
4/4 = 1.0
5/4 = 1.25
6/4 = 1.5
etc...

I highlighted the word integer because obviously the answer would be different if this wasn't stipulated by the question.

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by Ian Stewart » Mon Jul 07, 2008 9:07 am
lion's solution above is a very good one- you can see by choosing s = 3 for the first statement that it's not sufficient. And if the denominator is 4, there are only four possibilities: the decimal must end in .25, .5, .75 or the result will be an integer.

In general, for terminating decimal questions, you want to:

1) reduce fractions completely
2) prime factorize the denominator
3) if the denominator has a prime factor besides 2 or 5, the fraction represents a 'recurring' (non-terminating) decimal
4) if the only primes that divide the denominator are 2 and/or 5, the fraction represents a terminating decimal.

So, if 4 = 2^2 is the denominator, the fraction definitely terminates. Statement 1) only tells us about the numerator; that doesn't help us to know if the decimal terminates.
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by albertrahul » Tue Jul 08, 2008 2:33 am
A BIG thank you to both of you.
These explanations perfectly resolves my doubt :)