merajul wrote:
Hence n will not be divisible by 6. So Sufficient.
IMO, B
Hi,
I'm trying to understand this question too. The question I saw had n=K(K+7), not k+ 7
K as written in the question above. Can anyone explain this q with the change please?
Thanks![/quote]
If k is a positive integer and n = k(k + 7), is n divisible by 6?(1) k is odd. If \(k = 1\), then \(n = k(k + 7) = 8\) and n is NOT divisible by 6 but if \(k = 3\), then \(n = k(k + 7) = 30\) and n IS divisible by 6. Not sufficient.
(2) When k is divided by 3, the remainder is 2 --> \(k = 3x + 2\) --> \(n = k(k + 7) = (3x + 2)(3x + 9)=9x^2+33 x+18=3(3x^2+11x)+18\). Notice that \(3x^2+11x\) is even no matter whether x is even or odd, thus \(n=3(3x^2+11x)+18=3*even+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)\). Sufficient.
Answer: B.
Hope it's clear.[/quote]
(1) Since k=odd, putting k = 3,5,7 we get n=30,60,98. 30&60 divisible but 98 is not divisible by 6. So INSUFFICIENT.
(2) says, k=3a+2, putting value of k in n = k(k+7), we get
n=(3a+2) * (3a+9) = 9a2+27a+6a+18
Now, (〖9a〗^2+27a+6a+18)/6 => 〖9a〗^2/6 + 27a/6 + 6a/6 + 18/6 => 〖9a〗^2/6 + 27a/6 + 0 + 0
When.
a = even, Reminder is 0
a = odd, Reminder is 1+1=2 [ a = 3,5,7 etc.]
Combining (1) & (2) , Reminder will be 2 as shown above. So Answer Choice should be ‘C’ .
Bunuel]
Bunuel[/url] Can you please tell, where am I getting wrong ?[/quote]
Hi
merajul,
I will help you from the step you were stuck at that is statement (II)
You derived that n= n=(3a+2) * (3a+9) = 9\(a^{2}\)+27a+6a+18
Now lets solve this further
n= 9\(a^{2}\)+33a+18
n= 3(3\(a^{2}\)+11a) +18
for n to be divisible by 6 the part 3(3\(a^{2}\)+11a) need to be divisible by 6, because 18 is divisible by 6
Since we can see that 3(3\(a^{2}\)+11a) is divisible by 3 we need to know if its also divisible by 2
Now lets if that heppens
To determine if its divisble by 2 we need to know if its even or odd
Say a is even
then we have 3(Even+even)= 3(Even) so divisible by 6
say a is off then we have 3(Odd+Odd)= 3(even ) which is divisible by 6
So Statement (II) is sufficient .
Hope this helps
_________________
Probus
~You Just Can't beat the person who never gives up~ Babe Ruth