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If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even

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If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 09 Jul 2016, 10:40
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If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even?

(1) a>c
(2) b=c+d
[Reveal] Spoiler: OA

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 09 Jul 2016, 11:02
(2^a·3^b)/(2^c·3^d) = 2^(a - c) * 3^(b - d)
For 2^(a - c) * 3^(b - d) to be even (a - c) > 0 and (b - d) >= 0

St1: a - c > 0. Not Sufficient as we do not know about (b - d)

St2: b - d = c --> 2^(a - c) * 3^c. Clearly insufficient

Combining St1 and St2, 2^(a - c) = even
3^c may be an integer or a fraction depending on the value of c.
Not Sufficient

Answer: E

What is the OA?

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 09 Jul 2016, 14:15
Vyshak wrote:
(2^a·3^b)/(2^c·3^d) = 2^(a - c) * 3^(b - d)
For 2^(a - c) * 3^(b - d) to be even (a - c) > 0 and (b - d) >= 0

St1: a - c > 0. Not Sufficient as we do not know about (b - d)

St2: b - d = c --> 2^(a - c) * 3^c. Clearly insufficient

Combining St1 and St2, 2^(a - c) = even
3^c may be an integer or a fraction depending on the value of c.
Not Sufficient

Answer: E

What is the OA?


Not sure but think it's C.

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 09 Jul 2016, 22:57
ha15 wrote:
Vyshak wrote:
(2^a·3^b)/(2^c·3^d) = 2^(a - c) * 3^(b - d)
For 2^(a - c) * 3^(b - d) to be even (a - c) > 0 and (b - d) >= 0

St1: a - c > 0. Not Sufficient as we do not know about (b - d)

St2: b - d = c --> 2^(a - c) * 3^c. Clearly insufficient

Combining St1 and St2, 2^(a - c) = even
3^c may be an integer or a fraction depending on the value of c.
Not Sufficient

Answer: E

What is the OA?


Not sure but think it's C.


its E

take cases a=2,b=2,c=1, d=1 or a=5,b=4,c=4,d=1

this will E

either statements are not suff, coz info about 1 variable is not given

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 10 Jul 2016, 06:57
My approach.
St.1
We know that 2^a/2^c = Integer if a-c is an integer. But we don't know anything about b,d.
St.2
We know that b>d but we don't know anything about a and c.

Combining st 1 and st 2 we get that a>c and b>d which would give us what we want.
A straight 2^x and a 3^y which would result in an even number.
But we are not given that x,y is an integer.
So E.
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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 12 Jul 2016, 05:26
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Take into account that \(0<c<a\) or \(c<a<0\). The base of 2 won't be impacted but the base of 3 might. If -ve, then it's a fraction; if +ve, then even indeed.
Therefore, Option E

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 27 Jul 2017, 06:14
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ha15 wrote:
If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even?

(1) a>c
(2) b=c+d


Note here that a question such as "Is x an even integer?" does not imply that x must be an integer and that all we need to answer is whether it is even or odd. We do need to consider the cases where x may not be an integer.
Here it is relevant because it is not sufficient to know that a is greater than c. What if b is less than d? In that case the expression may not have an integer value.

b = c + d
shows that b is less than d in case c is positive
but b is greater than d in case c is negative

So even with both statements, both cases are possible.

Say, a = 4, c = 2, b = 4, d = 2
(2^a·3^b)/(2^c·3^d) = (2^4·3^4)/(2^2·3^2) = 4*9 = 36 (even integer)

Say a = 4, c = -2, b = 2, d = 4
(2^a·3^b)/(2^c·3^d) = (2^4·3^2)/(2^(-2)·3^4) = 64/9 (not an integer)

Answer (E)
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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even [#permalink]

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New post 01 Aug 2017, 22:39
Vyshak wrote:
(2^a·3^b)/(2^c·3^d) = 2^(a - c) * 3^(b - d)
For 2^(a - c) * 3^(b - d) to be even (a - c) > 0 and (b - d) >= 0

St1: a - c > 0. Not Sufficient as we do not know about (b - d)

St2: b - d = c --> 2^(a - c) * 3^c. Clearly insufficient

Combining St1 and St2, 2^(a - c) = even
3^c may be an integer or a fraction depending on the value of c.
Not Sufficient

Answer: E

What is the OA?


a,b,c,d are integers. So, they can be negative too!
now, if a=0 and c= -3, according to statement 2, b-d will become negative. Hence, b>d will no longer be true.

Therefore E.
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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even   [#permalink] 01 Aug 2017, 22:39
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