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

Question

A/B is a fraction such that A and B are co-prime positive integers. What can be the value of B such that A/B is not a non-terminating decimal?

(A) 28
(B) 42
(C) 84
(D) 128
(E) More than two answers are correct

Zarrolou · Accepted Answer

For  \frac{A}{B}  to be a terminating decimal b must have the form  2^a*5^b .

Every option except 128 contains a  7 , and since A and B do not share any factor (are co-prime) 7 will not cancel out of the denominator.

So B can only be  128=2^7

Bunuel · Answer

A and B are co-prime means that they do not share any common factor but 1.

Thus in order A/B to be a terminating decimal B must be of the form 2^n*5^m. Only D fits.

Answer: D.

Bunuel · Answer

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

Im2bz2p345 · Answer

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

PathFinder007 · Answer

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

Bunuel · Answer

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.

manhasnoname · Answer

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.

Fdambro294 · Answer

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

oldmonkey · Answer

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.

Fdambro294 · Answer

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

VijayLisha · Answer

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?

Bunuel · Answer

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.

pudu · Answer

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

Bunuel · Answer

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.

Aritrakarak · Answer

But 42 also contains a 7

Bunuel · Answer

Yes, exactly, that’s the point being made. All the other options (28, 42, and 84) include 7 as a factor, so they cannot produce a terminating decimal when A and B are co-prime. Only 128 (which is 2^7) has no factor other than 2 and 5, so it’s the only valid choice.

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

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

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