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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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09 Jul 2016, 10:40
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41% (01:24) correct 59% (02:58) wrong based on 236 sessions

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If a, b, c and d are integers, is $$\frac{2^a*3^b}{2^c*3^d}$$ even?

(1) a > c
(2) b = c + d
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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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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

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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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

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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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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12 Jul 2016, 05:26
Top Contributor
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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27 Jul 2017, 06:14
1
3
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)

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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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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

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]

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07 Nov 2017, 14:03
$$(2 ^ a . 3 ^ b)/(2 ^ c . 3 ^ d)$$ = even?

rearranging, 2 ^ (a-c). 3 ^ ( b - d) is even?

question can be simplified as: a > c and more importantly, b > d, else, resulting number will be fraction

Statement 1:
a > c, but we don't know if b > d . insuff

Statement 2:
b = c + d
it may look like b > d, but if d = 4, c = -2, then b = 2, => b < d
also we dont know if a > c -> insuff

Statement 1 + 2
also does not help us know if b > d

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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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04 Mar 2018, 09:03
2^a.3^b/2^c.3^d = 2^(a-c)*3^(b-d)

Statement 1:

A>C

If both a and c positive then indeed a-c>0 , but we don't know anything about b-d, if b-d is less than 0 then even if a>c , it won't make it even.

Just to consider , if both a and c are negative, still a-c>0, but still we don't know about b-d. Hence insufficient

Statement 2: b=c+d
2^a-c*3^(c+d-d)
2^a-c*3^c

We don't have info about a-c therefore insufficient .

Combining two:

2^(a-c) is even
However we know that 3^(b-d) = 3^c but we don't know if c itself is negative or positive, as question stem says it to be an integer
(I made it wrong while attempting the question myself)

Hence Ans is 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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19 Jun 2018, 09:01
1) a > c
This prove that equation will have one two.
But it doesn't give any info about b and d

(2) b = c + d

No info about a, since a value decides whether number is even or not

Combining both we have
At least one two
But we don't know the value of c+d if they are positive then figure will give even
If it is negative then 3 will have negative sign and will be dividing ,we don't that value would come out to be even or odd so No

Multiple cases on combining

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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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19 Jun 2018, 09:16
Clearly both the statements alone are NS. If you combine them & then you test for positive & negative integers then you get different answers. Hence NS.

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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even &nbs [#permalink] 19 Jun 2018, 09:16
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