Which of the following has a decimal equivalent that is a

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 the 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 QUESTION:

I. 1/12 = 1/(2^2*3)

II. 1/10^2 = 1/(2*5)^12

III. 1/2^10

According to the above only II and III will be terminating decimals.

Answer: E.

Questions testing this concept:
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Hope it helps.
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When solving this problem, we should remember that there is a special property about fractions that allows their decimal equivalents to terminate. This property states:

“In its most-reduced form, any fraction with a denominator whose prime factorization contains only 2s, only 5s, or both 2s and 5s, produces decimals that terminate. A denominator with any other prime factors produces decimals that do not terminate.”

Let’s consider each of the Roman numerals and determine which fractions have only 2s, 5s, or both as prime factors.

I. 1/12

Since 1/12 = 1/(3 x 2^2), 1/12 is NOT a terminating decimal.

II. 1/10^2

Since 1/10^2 = 1(2^2 x 5^2), 1/10^2 IS a terminating decimal.

III. 1/2^10

1/2^10 IS a terminating decimal.

Answer: E
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Hi Bunuel,

I used the theory from GMAT club but got the question wrong because i assumed you need both 2 and 5. Does the rule imply that any denominator with a 2 or a 5 and not necessarily both 2 and a 5(with a positive exponent) = terminating decimal ?
Also, can we conclude that only non-terminating decimals when another prime number is in the denominator other than 2 and 5(assuming the fraction is in its most reduced form of course).

Please read the theory bit carefully. It says "\(2^n5^m\), where m and n are non-negative integers". So, n or m could be 0. This means that the denominator must have either only 2's, only 5's or any combination of 2's and 5's.

For similiar questions check Trailing Zeros Questions in our Special Questions Directory.
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I also used your approach. First I also thought that I would need 5 as well, but fortunately I remembered that we could also have 5^0

however, I convinced myself by just randomly dividing 1/2, 1/4, 1/8 and so on till I was confident enough to move on =)

Inclusion of 3 in 12 in answer option A will lead to a non-terminating decimal.
Numbers 2 and 5 don't end in Non terminating decimals , hence answer option E.
Time taken : 4 seconds.

A terminating decimal would be any number divided by 2,5, or the combination of 2 and 5 (10).

1/2 = 0.5
1/5 = 0.2
1/10 = 0.1
1/50 = 0.02 etc..

Only II and III fit this property.

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