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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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09 Jul 2016, 10:40
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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
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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
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
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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
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
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10 Jul 2016, 06:57
My approach. St.1 We know that 2^a/2^c = Integer if ac 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
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12 Jul 2016, 05:26
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
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27 Jul 2017, 06:14
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
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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
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, bd 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
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07 Nov 2017, 14:03
\((2 ^ a . 3 ^ b)/(2 ^ c . 3 ^ d)\) = even?
rearranging, 2 ^ (ac). 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
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
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04 Mar 2018, 09:03
2^a.3^b/2^c.3^d = 2^(ac)*3^(bd) Statement 1: A>C If both a and c positive then indeed ac>0 , but we don't know anything about bd, if bd 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 ac>0, but still we don't know about bd. Hence insufficient Statement 2: b=c+d 2^ac*3^(c+dd) 2^ac*3^c We don't have info about ac therefore insufficient . Combining two: 2^(ac) is even However we know that 3^(bd) = 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 Posted from my mobile device



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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even
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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 So E is answer
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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even
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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. E is correct answer.
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Re: If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even &nbs
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19 Jun 2018, 09:16






