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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 09 Jan 2012, 21:58
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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
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B
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 13 Jan 2012, 16:36
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enigma123
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?
Show more

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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 28 Jul 2012, 09:10
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enigma123
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?
Show more

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....)
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 09 Jan 2012, 23:32
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 16 Jan 2012, 16:39
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Bunuel - you are a LEGEND. Many thanks for the lovely explanation.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 28 Jul 2012, 05:34
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@Bunuel

What if e=0 ? Will it be a terminating decimal ?
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 28 Jul 2012, 05:54
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@Bunuel

What if e=0 ? Will it be a terminating decimal ?
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 Updated on: 02 Sep 2012, 06:29
Originally posted by Zinsch123 on 02 Sep 2012, 06:20.
Last edited by Zinsch123 on 02 Sep 2012, 06:29, edited 2 times in total.
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 04 Nov 2012, 23:02
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 05 Nov 2012, 11:44
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himanshuhpr
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.
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 25 Oct 2014, 23:58
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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
enigma123
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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 26 Oct 2014, 06:32
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TARGET730
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
enigma123
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.
Show more

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.

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 06 Dec 2015, 11:31
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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.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 21 Sep 2017, 11:45
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Bunuel
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This question needs some editing, i first read the question as: p = \(2^{(3a)^{b}\) and q = \(2^{(3c)^{(5d)^{e}}\)

So I answer D because the denominator consists of 2 only so the decimal will terminate regardless :lol:
_______________
Edited. Thank you.
Show more

Hi Bunuel,

I have a doubt. If denominator has only 2 or 5 with any power will terminate irrespective of numerator (I mean whether numerator is integer or not?) If so can please explain why answer is C for below question? In explanation it was mentioned that if P=1/6 then result will be something. But my question is p is a factor of 100.. So how can one assume 1/6 as value of p .. Aren't factors always positive integers?

I tried to find this question on gmatclub but could get any information. If it is repeating I will correct my mistake.

Is P/Q a terminating decimal?
(1) P is a factor of 100 (2) Q is a factor of 100
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 21 Sep 2017, 12:04
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Bunuel
NamVu1990
This question needs some editing, i first read the question as: p = \(2^{(3a)^{b}\) and q = \(2^{(3c)^{(5d)^{e}}\)

So I answer D because the denominator consists of 2 only so the decimal will terminate regardless :lol:
_______________
Edited. Thank you.

Hi Bunuel,

I have a doubt. If denominator has only 2 or 5 with any power will terminate irrespective of numerator (I mean whether numerator is integer or not?) If so can please explain why answer is C for below question? In explanation it was mentioned that if P=1/6 then result will be something. But my question is p is a factor of 100.. So how can one assume 1/6 as value of p .. Aren't factors always positive integers?

I tried to find this question on gmatclub but could get any information. If it is repeating I will correct my mistake.

Is P/Q a terminating decimal?
(1) P is a factor of 100 (2) Q is a factor of 100
Show more

1. When it's mentioned "reduced fraction a/b", it's implied that both a and b are integers.
2. Only positive integers can be factors on the GMAT, so I don't know why the solution you mention considers 1/6 as a factor. I would not trust such source.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 21 Sep 2017, 20:35
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Bunuel
goalMBA1990

Thank you very much for clarification. So answer we would be B for above question. Correct?

Sent from my XT1663 using GMAT Club Forum mobile app

In the below question:
Is P/Q a terminating decimal?

(1) P is a factor of 100
(2) Q is a factor of 100

Does the solution mention that p can be 1/6 for the first statement or for the second? If for (2), then yes p could be 1/6 and in this case (2) is not sufficient. When combined, since p is a factor of 100, then it must be an integer, thus p/q will be terminating decimal and the answer would be C.

Whether it is mentioned or not, how any fraction become factor of any integer? How can we assume fraction for P or Q?

Sent from my XT1663 using GMAT Club Forum mobile app
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For (1) since p is a factor of 100, then it must be an integer. But q can be any real number: fraction, integer, irrational, ...
For (2) since q is a factor of 100, then it must be an integer. But p can be any real number: fraction, integer, irrational, ...
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 23 Oct 2017, 23:19
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Bunuel
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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.
Show more



-----------------

as you said that both the numerator and denominator should be reduced to the lower term ,,, but as in the question, it was clearly seen that 2^a and 2^c is present in both numerator and denominator so the term is not reduced to its lowest term...

now from statement 1, we can clearly state that 2 will not present in the denominator and we have to have only 2 and 5 in the denominator to say that the term would be terminating. Hence statement 1 looks sufficient to me

could you please help me explain this ?
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 24 Oct 2017, 00:06
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priyanknema
Bunuel
enigma123
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.



-----------------

as you said that both the numerator and denominator should be reduced to the lower term ,,, but as in the question, it was clearly seen that 2^a and 2^c is present in both numerator and denominator so the term is not reduced to its lowest term...

now from statement 1, we can clearly state that 2 will not present in the denominator and we have to have only 2 and 5 in the denominator to say that the term would be terminating. Hence statement 1 looks sufficient to me

could you please help me explain this ?
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If the denominator of a reduced fraction has only 2's OR only 5's OR only 2's and 5's the fraction will be a terminating decimal.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 15 Apr 2018, 21:50
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Bunuel
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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.
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Hi Bunuel,

what if c and e are negative?
Since the questions says they are integers so they could be negative as well.
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If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 15 Apr 2018, 22:46
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vedanshimurarka
Bunuel
enigma123
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.


Hi Bunuel,

what if c and e are negative?
Since the questions says they are integers so they could be negative as well.
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Hello

Good point I think. But then again if the powers can be negative integers, we should only be concerned with 'b' and 'd' here (since a, c, e are powers of 2's/5's and whether their powers are positive or negative doesn't matter in this question as with any integer powers of 2's/5's it will still remain a terminating decimal. So our concern is with powers of 3 here, because after cancelling all terms, after reducing this fraction to its lowest form, if there still remains a '3' or a higher power of '3' in the denominator then it will turn out to be a non-terminating decimal, but if '3' is completely removed from the denominator then it will be a terminating decimal).

So now even if b and d are say negative integers (or one of them is negative and other is non-negative):- still second statement tells us that b > d. Since b > d, in this case we will have 3^(b-d) in the numerator. No matter what signs b/d have, if b > d, then b-d will be > 0 and thus we will have 3^(positive integer) in the numerator. This means 3 will be completely eliminated from the denominator, and thus this will be a terminating decimal indeed. Hence B answer.

(You can try taking negative values of b/d but where b > d, you will find that there will always be a terminating decimal as 3 will be removed from the denominator)
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