Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Terminating Decimal [#permalink]
09 Jan 2012, 22:32

As I understand, in order to be a non-terminating decimal we should be able to convert a number into X/99 format. If b>d then there is no way we can get 99 in the denominator and hence it will always be a terminating decimal. Thus, B is an answer.

Re: Terminating Decimal [#permalink]
13 Jan 2012, 15:36

8

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

enigma123 wrote:

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? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

Theory: Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and onlyb (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 ORIGINAL QUESTION: 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?

Question: is \frac{2^a*3^b}{2^c*3^d*5^e} a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d?

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]
28 Jul 2012, 04:54

Expert's post

smartmanav wrote:

@Bunuel

What if e=0 ? Will it be a terminating decimal ?

You mean for (2)? In this case the denominator will have only 2's in it, and if the denominator has only 2's or only 5's in it, it still will be a terminating decimal. _________________

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]
28 Jul 2012, 08:10

enigma123 wrote:

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? (1) a > c (2) b > d

Any idea what is the concept behind this question to get a answer B?

Of all the theories.

Among 1/2, 1/3 and 1/5, only 1/3 is non terminating. So if we don't have 3 in the denominator then only p/q will be terminating. b>d, ensures we have no "3" left in the denominator, hence the decimal is terminating. (it holds true for 7,11,13....) _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Re: is p/q a terminating decimal? [#permalink]
02 Sep 2012, 05:20

1

This post received KUDOS

The question here is, whether b >= d. Why is that? p and q are given in their prime factorization. If q has more twos and/or fives in its prime factorisation than p, it won't result in a non-terminating decimal, Remainder of 2 can only be 1: 1/2=0.5 and remainders of 5 result in: 1/5=0.2, 2/5=0.4 3/5=0.6 and 4/5=0.8.

However, this is not the case with the 3. If q has more threes than p, you can cancel all of the threes in the numerator, but there will remain some threes in the denominator, resulting in a non-terminating decimal, because 1/3=0.33333 and 2/3=0.666666

Statement (1) gives us no information about b and d. Statement (2) does. There are fewer threes in the denominator. They will cancel with some of the threes in the numerator. Therefore, this statement is sufficient. We know that p/q will be a terminating decimal.

I hope my explanation is good enough.

Last edited by Zinsch123 on 02 Sep 2012, 05:29, edited 2 times in total.

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]
05 Nov 2012, 10:44

2

This post received KUDOS

himanshuhpr wrote:

what if b = -2 & d = -3 , then we have a case for terminating decimal ?? because the denominator now would be in 2^m * 5^n form.

Yes, p/q will be a terminating decimal. For b = -2 and d = -3, b > d.

Since p/q = 2^{a-c}3^{b-d}5^{-e}, the given ratio is a terminating decimal if and only if b-d\geq{0} or b\geq{d}. Which means there is no factor of 3 in the denominator, only factors of 2 and/or 5, if at all. If in addition a\geq{c} and e\leq{0}, the given ratio is in fact an integer, which is a terminating decimal. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]
06 Dec 2013, 03:00

Thanks Bunuel, but I think I worded my question poorly.

My questions is stemming from if whether A could equal 0, and c>0, which would make it 2^0 which would make it equal to 1. so the denominator would be 2^03^d5^e . _________________

4/28 GMATPrep 42Q 36V 640

gmatclubot

Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
[#permalink]
06 Dec 2013, 03:00