A/B is a fraction such that A and B are co-prime positive

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 \(2*5^2\). 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.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

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The question does not ask to find the value of B. It asks: "What CAN be the value of B such that A/B is not a non-terminating decimal?" Though B can take infinitely many values, from the options provided it can only be 128.

Small correction to the above theory - should be 5^3 (not 5^2).

Thanks for the in-depth post and simple to understand explanation Bunuel!

~ Im2bz2p345

Hi Bunuel,

I have a doubt here question is asking A/B to be non terminating decimal here we are saying A/B to be terminating decimal it should be of form 2^n5^m pleae clarify.

Thanks

Please read the question carefully: what can be the value of B such that A/B is not a non-terminating decimal? So, the question asks for such value of B so that A/B will be terminating decimal.

In simpler words, if B has only 2s and or 5s and nothing else in its prime factorization, then A/B will not be a non-terminating decimal.

clever wordplay with the double negative ---- "... NOT a NON-terminating decimal"

---in other words, IS a Terminating Decimal


if the NUM and DEN are Co-Prime, they do NOT Share Any Prime Bases and the Fraction can not be simplified any further.

Thus, we are looking for an Answer the provides a DEN that has ONLY Prime Bases of: 2 and/or 5


128 = (2)^7

All the other Answers are NOT of the Form: (2)^n * (5)^m ---- in which Exponents n and m = NON-Negative Integers


D

The question doesn't make it clear that the value of B has to be from the provided options. I went for E and then realized from the answer explanation that it was to be assumed that you have to limit the answer to provided choices. Any co-prime with 2 & 5 as B would provide you a terminating decimal.

I see what you’re saying. I think it’s good that a GMAT quant question will never have this kind of ambiguity. I’ve never seen an answer choice worded in such a way as “more than 2 answer are correct.”

Posted from my mobile device

Hello Buenel,
I'm having doubt related to co prime. Question stem is saying a & b are co prime. But we are doing checks whether B contains 2^x *5^y alone. Would you explain a little bit?

I believe this is already explained above. However, here it goes again:

A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) 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 the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of only 2s and/or 5s, 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 a terminating decimal.

(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)

In the original question, we are given that a and b are co-prime, which implies that the fraction \(\frac{a}{b}\) is already in its simplest form, so reduced to its lowest term. Therefore, no \(\frac{a}{b}\) where b is not in the form of \(2^n5^m\) could be terminating.

still i didn't understand. 128/5^n2^0 can be terminating. similarly 28/5^n can be terminating. I meant 7/5, 7/25/7/125 etc.

then how we decided it is only 128. please help me