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Can some one tell how to solve it in a faster way?

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 \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

\(\frac{39}{128}=\frac{39}{2^7}\), denominator has only prime factor 2 in its prime factorization, hence this fraction will be terminating decimal.

All other fractions (after reducing, if possible) have primes other than 2 and 5 in its prime factorization, hence they will be repeated decimals.

numbers with terminating decimals basically should have 5 or 2 or both in its denominators, right? So any numerator with denominator 125 or 8 would be a terminating decimal?

numbers with terminating decimals basically should have 5 or 2 or both in its denominators, right? So any numerator with denominator 125 or 8 would be a terminating decimal?

Thanks.

Yes, as denominator 125=5^3 or 8=2^3, numerator can be any integer.
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Hi Bunuel As per your explanation if the denominator is not in the form of 2^n 5^m then the fraction will be terminal decimal. If you look at the denominator of other answer choices they are also not in the above form 1. 189 = 3^3 *7^1 2. 196 = 2^2 * 7^2 3. 225 = 3^2 * 5^2 4. 144 = 2^4 * 3^2

So how the last answer choice is correct still not clear based on your explanation?

Hi Bunuel As per your explanation if the denominator is not in the form of 2^n 5^m then the fraction will be terminal decimal. If you look at the denominator of other answer choices they are also not in the above form 1. 189 = 3^3 *7^1 2. 196 = 2^2 * 7^2 3. 225 = 3^2 * 5^2 4. 144 = 2^4 * 3^2

So how the last answer choice is correct still not clear based on your explanation?

As per solution:

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 \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

A. \(\frac{10}{189}=\frac{10}{3^3*7}\) --> denominator has primes other than 2 and 5 in its prime factorization, hence it's repeated decimal;

B. \(\frac{15}{196}=\frac{15}{2^2*7^2}\) --> denominator has primes other than 2 and 5 in its prime factorization, hence it's repeated decimal;

C. \(\frac{16}{225}=\frac{16}{3^2*5^2}\) --> denominator has primes other than 2 and 5 in its prime factorization, hence it's repeated decimal;

D. \(\frac{25}{144}=\frac{25}{2^4*3^2}\) --> denominator has primes other than 2 and 5 in its prime factorization, hence it's repeated decimal.

E. \(\frac{39}{128}=\frac{39}{2^7}\), denominator has only prime factor 2 in its prime factorization, hence this fraction will be terminating decimal. All other fractions' denominator have primes other than 2 and 5 in its prime factorization, hence they WILL BE repeated decimals:

Re: Which of the following fractions has a decimal equivalent [#permalink]

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05 Dec 2013, 11:59

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Re: Which of the following fractions has a decimal equivalent [#permalink]

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10 Dec 2013, 19:23

First thought: Terminatieng and non terminating My concept after reading the question: Terinating – non repeating, Non terminating – Repeating numbers after decimal My Strategy : 1.All numbers are squares or cubes 2. Simply these 3. Then divide I have learned – Denominator having 2^m5^n are terminating numbers
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I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

First thought: Terminatieng and non terminating My concept after reading the question: Terinating – non repeating, Non terminating – Repeating numbers after decimal My Strategy : 1.All numbers are squares or cubes 2. Simply these 3. Then divide I have learned – Denominator having 2^m5^n are terminating numbers

Dear kanusha, I am responding to your private message.

What you say in the last line is correct and is key to understanding this problem. My only caution would be: use proper mathematical grouping symbols. You are not thinking like a mathematician when you write 2^m5^n That is precisely the way it is written by someone who isn't thinking carefully about the mathematical symbols. What you meant is: (2^m)(5^n) Those parenthesis are not garnish, not extra decorative elements --- they are absolute essential pieces of mathematical equipment, and you are setting yourself up for mistake if you casually ignore their tremendous importance. See: http://magoosh.com/gmat/2013/gmat-quant ... g-symbols/

In this problem, all of the denominators happen to be squares and other powers. 289 = 17^2 196 = 14^2 225 = 15^2 144 = 12^2 128 = 2^7 I think it's good to know the perfect squares up to 20^2 = 400. It's also good to know the first eight powers of 2. It just saves time, and helps to deepen number sense.

Nevertheless, the fact that most of these are squares is not particularly relevant. All you have to do is find the prime factorization of the denominator. As soon as you find a prime factor other than 2 or 5, then you know the decimal would be repeating & non-terminating.

If the denominator an odd number not ending in a 5, then it can't be divisible by 2 or 5: it must have other prime factors and must lead to a repeating & non-terminating decimal. If the denominator is divisible by 3, a very easy check, then it lead to a repeating & non-terminating decimal.

The easy way to handle these, even without knowing they are perfect squares ---- (A) 289 --- an odd number, so not divisible by 2, and clearly not divisible by 5, so it must have other prime factors. No good. (B) 196 --- divide by 2 = 98 --- divide by 4 = 49 --- other odd factors. No good. (C) 225 --- 2 + 2 + 5 = 9, which is divisible by 3, so that means 225 is divisible by 3. No good. (D) 144--- 1 + 4 + 4 = 9, which is divisible by 3, so that means 144 is divisible by 3. No good. (E) only one left

So, as that blog explains, if the denominator of a fraction has no prime factors other than 2 and 5, the fraction will terminate instead of repeat.

The next step is to recognize that 128 is a power of 2. It's highly worthwhile to have the first ten powers of 2 memorized: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 2^9 = 512 2^10 = 1024

Since 128 is a power of 2, it has only factors of 2, no other prime factors. This means, any fraction with 128 in the denominator will be a terminating decimal.

If you have any questions after you read that blog post, please let me know.

Re: Which of the following fractions has a decimal equivalent [#permalink]

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09 Jul 2015, 11:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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