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For a fraction to yield a terminating decimal, its denominator must be composed ONLY of powers of 2 and/or of powers of 5 when the fraction is in its MOST REDUCED FORM.

In any multiple of 3, the sum of the digits is a multiple of 3.
Thus, the denominators of A (189), C (225), and D (144) are all multiples of 3.
None of these answer choices can be further reduced.
Since each of these answer choices is in its most reduced form, and each has in its denominator something other than a power of 2 or a power of 5, none will yield a terminating decimal.
Eliminate A, C and D.

Answer choice B: 15/196 = (3*5)/(2*2*7*7)
Since in its most reduced form this fraction has in its denominator something other than a power of 2 or a power of 5, the resulting decimal will not be terminating.
Eliminate B.

The correct answer is E.

Answer choice E: 39/128 = (3*13)/(2^7).
Since in its most reduced form this fraction has in its denominator only a power of 2, the resulting decimal will be terminating.

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Here is a shortcut method: A fraction in lowest terms with a prime denominator other than 2 or 5 always produces a repeating decimal.

Thus if the denominator of a fraction after canceling all the common factors with the numerator contains any prime other than 2 or 5 in their prime factorization, then the fraction does not have a decimal equivalent that is a terminating decimal.

A] 10/189 --> 189 contains 3, 10 doesn't => Repeating Decimal
B] 15/196 --> 196 contains 7, 15 doesn't => Repeating Decimal
C] 16/225 --> 225 contains 3, 16 doesn't => Repeating Decimal
D] 25/144 --> 144 contains 3, 25 doesn't => Repeating Decimal
E] 39/128 --> 128 contains 2 only => Terminating Decimal


The correct answer is E.

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Anurag Mairal, Ph.D., MBA
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