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31 Jan 2018, 06:01
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C
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Difficulty:
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Question Stats:
59% (02:30) correct
41%
(02:19)
wrong
based on 191
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If a and b are positive integers, is \(a^3 + b^2 + 4\) divisible by 6?
(1) b is odd
(2) \(\frac{5a}{b}\) is even
(1) b is odd
(2) \(\frac{5a}{b}\) is even
Mo2men
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31 Jan 2018, 06:36
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Is \(a^3 + b^2 + 4\) divisible by 6?
(I) b is odd
No info about a
Let's take exmaple
a =1, b= 1...........\(\frac{6}{6}\) = 1 ............Answer to question is Yes
a =2, b= 1...........\(\frac{13}{6}\) is Not integer ............Answer to question is No
Insufficient
(II) \(\frac{5a}{b}\) is even
Let's analyze first:
Case 1:
\(\frac{even}{even}\) = even
This means a = even & b =even.......This means we can have either number divisible by 6
Let a=4, b =10.......\(\frac{20}{10}\)=2... Apply in question, we get 168, which is even and its sum is divisible by 3....so it is divisible by 6......Answer is Yes
We can choose a = 8 and b =10........Answer is No
I could say insufficient. But I want to examine case 2
Case 2:
This means a = even & b =odd ( for example a = 2, b =5)
This means directly.......(even)^3 + (odd)^2 + 4 =odd which can't be divisible by 6....Answer is No (please note that exponent does NOT change nature of integer if odd or even)
combining 1 & 2
It is clear we have case 2 ans straight forward answer will be always NO
Sufficuent
Answer: C
(I) b is odd
No info about a
Let's take exmaple
a =1, b= 1...........\(\frac{6}{6}\) = 1 ............Answer to question is Yes
a =2, b= 1...........\(\frac{13}{6}\) is Not integer ............Answer to question is No
Insufficient
(II) \(\frac{5a}{b}\) is even
Let's analyze first:
Case 1:
\(\frac{even}{even}\) = even
This means a = even & b =even.......This means we can have either number divisible by 6
Let a=4, b =10.......\(\frac{20}{10}\)=2... Apply in question, we get 168, which is even and its sum is divisible by 3....so it is divisible by 6......Answer is Yes
We can choose a = 8 and b =10........Answer is No
I could say insufficient. But I want to examine case 2
Case 2:
This means a = even & b =odd ( for example a = 2, b =5)
This means directly.......(even)^3 + (odd)^2 + 4 =odd which can't be divisible by 6....Answer is No (please note that exponent does NOT change nature of integer if odd or even)
combining 1 & 2
It is clear we have case 2 ans straight forward answer will be always NO
Sufficuent
Answer: C
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31 Jan 2018, 07:15
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EgmatQuantExpert
Statement I: Take \(a =1, b = 1\)
\(a^3 + b^2 + 4\) is divisible by 6.
Take \(a = 2, b = 1\)
\(a^3 + b^2 + 4\) is not divisible by 6. So, Insufficient.
Statement II:
Take \(a = 0, b =\) \(\sqrt{2}\)
\(a^3 + b^2 + 4\) is divisible by 6.
Take \(a = 1, b = 1/2\)
\(a^3 + b^2 + 4\) is not divisible by 6.
By Combining both we get,
\(b = odd, a =even\)
So, Option C.
