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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.

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 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 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]

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28 Jul 2012, 05:54

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]

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28 Jul 2012, 09:10

1

This post received KUDOS

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!

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, 06:29, 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]

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05 Nov 2012, 11:44

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]

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24 Oct 2014, 08:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

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25 Oct 2014, 23:58

Hi Bunuel,

Quick question on this rule. How about 1/15? it can be written as 1/2^0 * 3 * 5. The denominator has 5, but the fraction is not a terminating decimal. Can you please explain why?

Bunuel wrote:

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 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 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?

If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 [#permalink]

Show Tags

26 Oct 2014, 06:32

Expert's post

TARGET730 wrote:

Hi Bunuel,

Quick question on this rule. How about 1/15? it can be written as 1/2^0 * 3 * 5. The denominator has 5, but the fraction is not a terminating decimal. Can you please explain why?

Bunuel wrote:

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 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 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?

(1) a > c. Not sufficient. (2) b > d. Sufficient.

Answer: B.

Hope it helps.

1/15 = 1/(3*5). For a reduced fraction to be terminating, the denominator of the fraction should NOT have any prime but 2 or/and 5.

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

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05 Dec 2015, 08:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

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06 Dec 2015, 11:31

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

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

We can derive from p/q=2^a3^b/2^c3^d5^e, that b>=d as the denominator has to be of only 2 or 5 out of the prime factors, so 3 is eliminated and (B) hence becomes the answer.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

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