Which of the following fractions has a decimal equivalent that is a te

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Hope it helps.

Bunuel:

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.

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:

Hope it's clear.

Answer: E.

You can solve this question in less than 30 seconds if you understand the concept of terminating decimal. The denominator must have only power's of 2 or 5 in the denominator no other powers ( if it has any other prime factors like 3,7, etc it won't be terminating). 2's and 5's can be in any possible combination but it must only have 2's and 5's

a) 10/189

denominator sum of digits is 18 so its divisible by 3 eliminate

b) 15/196

This has a prime factor of 7 when do the prime factorization of the denominator.. Eliminate

c) 16/225

denominator sum of digits is 9 so its divisible by 3 eliminate

d) 25/144

denominator sum of digits is 9 so its divisible by 3 eliminate

e) 39/128 Jackpot Correct answer

This is true if a fraction is reduced to its lowest term.

Consider this: the denominator of 3/30 has other primes than 2 or 5, but 3/30 IS a terminating decimal because 3 in the denominator gets reduced: 3/30=1/10=0.1.

Hope it's clear.

I'm confused, 128 is 2^7, you said it had to be 2^n*5^m....there is no 5^m in 128

Yes, it is 128 = 2^7*5^0.

So ANY number with a 2^x or 5^x (where x is greater than or equal to 1) will fall into this then?

I'm confused though, so 6/15 a terminating decimal, because 15 is 5^1*3^1*2^0, but then why is 16/225 is not terminating? It follows the same pattern; 225 is 5^2*3^2*2^0

Please read again: which-of-the-following-fractions-has-a-decimal-equivalent-159322.html#p1264673

6/15=6/(3*5) is a terminating decimal because extra 3 in the denominator is reduced and we get 2/5 (the denominator is in the form of 2^n*5^m).

16/225=16/(3^2*5^2) is not a terminating decimal because extra 3^2 in the denominator is not reduced to get the denominator in the form of 2^n*5^m.

Hope it's clear.

Dear kanusha,
I am responding to your private message.

First of all, you may find this blog helpful.
https://magoosh.com/gmat/2012/gmat-math- ... -decimals/

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:
https://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

Does all this make sense?
Mike

Dear uwengdori,
I'm happy to respond. You may find some help in the other posts in this merged thread, but I will be happy to explain it as well.

First of all, I would highly suggest reading this post, which will clarify a great deal:
https://magoosh.com/gmat/2012/gmat-math- ... -decimals/

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.

Mike

NOTE: this is one of those questions that require us to check/test each answer choice.
In these situations, always check the answer choices from E to A, because the correct answer is typically closer to the bottom than to the top.

For more on this strategy, see my article: https://www.gmatprepnow.com/articles/han ... -questions

Cheers,
Brent

Check out this post for a detailed discussion on terminating decimals:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... -the-gmat/

A decimal will terminate when the equivalent fraction in lowest terms has a denominator that breaks down to primes of 2’s only, 5’s only, or both 2’s and 5’s only.

Looking at our answer choices we see that 39/128 is a terminating decimal since 128 = 2^7.

Answer: E

Bunuel, this is a GMAT Prep question. Could you please tag it as such?

Thank you!