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If k is a positive integer and n = k(k + 7k), is n divisible

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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 29 Mar 2016, 22:34
1
iliavko wrote:
VeritasPrepKarishma

Hello Karishma! A question here.. how do you quickly pull the 3 out of the initial statement?
(3b + 2)(3b + 9) = this one-> 3*(3b + 2)(b + 3)

Thank you!


Both terms of (3b + 9) have 3 as a factor.
(3*b + 3*3)

So you pull out a 3 from the brackets.
3*(b + 3)

Initial expression becomes
(3b + 2) * 3 * (b + 3)

The three terms are multiplied so you can arrange them in any way you like.

3 * (3b + 2) * (b + 3)
or
(b + 3) * (3b + 2) * 3
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 16 Jun 2016, 14:09
sandeepmanocha wrote:
Jem2905 wrote:
If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.



Divisibility by 6 means must be divisible by 3x2 (Prime Factors of 6)

1) k is Odd
==> k(k+7) = Odd(Odd+Odd) = Even
This will give me at least one factor of 2.. but nothing about factor of 3

[Insufficient]

2) k Divided by 3 ==> Remainder = 2
Now Possible values could be
2,5,8,11 -- Alternatively Even and Odd
So if k=2,8,...
Then I am getting One factor of 2 and when I add these to 7 I get Multiple of 3,

Similarly when I I choose k=5,11.. Odd Number with remainder of 2 and I add them to 7 I get Even Multiple of 3 i.e. Multiple of 6

This is happening because, Numbers with Remainders of 2 when added to 7, which leaves remainder of 1 when divided by 3, makes the Sum Divisible by 3
[Sufficient] <--- Answer is B



Agreed with Sandeep, if K is divisble by 3 and the remainder is 2, then (k+7) would divide evenly by 3 as 7/3 has a remainder of 1.
So 7/3=2*3+1 & k/3=a*3+2 -> (k+7)/3=(a*3+2)+(2*3+1)=3a+3+2*3=3(a+1+2) ->this is divisble by 3
Now, if k is odd k(k+7) is even, and if k is even k(k+7) is even, which means it is divisible by 2. Therefore it is divisible by both 2 and 3.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 29 Jan 2017, 16:34
1) First statement is insufficient. Smart numbers: k=1: 1*8=8, 8/6=1.666; k=3: 3*10=30, 30/6=5
2) To find out if n is divisible by 6 we need to find if n is divisible by 2 and divisible by 3. If k is even (e.g. 8, 14, 20...) then it is divisible by 2. If k is odd (5, 11, 17...), then k(odd)+7(odd)=even and it is divisible by 2. Since k/3 has a remainder of 2, and 7/3 has a remainder of 1, the remainders add up to another 3, thus k+7 is always divisible by 3.

Statement 2 alone is sufficient, while statement 1 alone is not sufficient.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 02 Jul 2017, 10:15
1
If n=K(K+7), then-
1) k is odd => odd * (odd + Even) = Odd * Odd. We can not say it is multiple of 6 or not. Insufficient.
2) k = 3m +2. So, n=K(K+7) => n= (3m+2)(3m+9)= 3(3m+2)(m+3) => multiple of 3.
If m is odd, m+3 is even. Hence , multiple of 2 and 3. So, it is multiple of 6. Suff.
If m is even, 3m+2 is even. Hence , multiple of 2 and 3 again. So,multiple of 6. Suff.
Ans- B is
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 22 Jul 2017, 19:01
Jem2905 wrote:
If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.



The goal is to find if n is divisible by 2 and 3 (6).

Statement 1) K is odd.

K = 1
1(1+7) = 8, not divisible by 6

K = 3
3(3+7) = 30, divisible by 6.

Not sufficient.

Statement 2) K/3 R = 2.

If k/3 has a remainder of 2, then k could be equal to the following integers:

2 ==> 2(2+7) = 18, divisible by 6.

5 ==> 5(5+7) = 60, divisible by 6.

8 ==> 8(8+7) = 8*15 = 80+40 = 120, divisible by 6.

Sufficient.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 26 Sep 2017, 00:30
Got it in 5 minutes :(.
either way, St 1: n = k(k+7)means; plug in numbers.
if k is odd; then odd (odd + odd) --> odd.
if k is even; then even (even + odd) --> even.
Not suff

St 2: k/3 = Q + 2
5/3 = 1 + 2
8/3 = 2+ 2
So the lowest number completely divisible would be 5 (numerator: 3+2 and deno : 3+2). So 5/5 would be 1
Plugging K back to 5 --> 5(12) = 60 which is divisible by 6.
Thus B is sufficient.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 05 Oct 2017, 07:12
VeritasPrepKarishma wrote:
Jem2905 wrote:
If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.


Given: n = k(k + 7)
Question: Is n divisible by 6?

(1) k is odd.
If k = 1, n = 8 - Not divisible by 6
If k = 6, n is divisible by 6
Not sufficient

(2) When k is divided by 3, the remainder is 2.
k = (3b+2)
n = (3b+2)(3b+2 + 7) = (3b + 2)(3b + 9) = 3*(3b + 2)(b + 3)
For n to be divisible by 6, it must be divisible by both 2 and 3. We see that it is divisible by 3. Let's see if it is divisible by 2 too i.e. if it is even.
b can be odd or even in this expression. If it is odd, (b+3) will become even because (Odd + Odd = Even). If it is even, (3b+2) will become even because (Even + Even = Even). So in either case, n will be even. So n will be divisible by 3 as well as 2 i.e. it will be divisible by 6.
Sufficient alone.

Answer (B)

I suppose what I have marked in Red is a typo since the statement says k is odd.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 21 Oct 2017, 09:05
Hence n will not be divisible by 6. So Sufficient.

IMO, B[/quote]

Hi,
I'm trying to understand this question too. The question I saw had n=K(K+7), not k+ 7K 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 Can you please tell, where am I getting wrong ?
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 24 Oct 2017, 08:58
I timed my response to work the problem--while knowing the answer--from the simplest solution I could find (quoted below) and it still took me ~3:30 mins/secs.

Any thoughts on how to cut down on the time used to solve? Has anyone else measured their response time?


KamiG wrote:
1. K is odd

It's clearly insufficient. Test k=1 and k=3.

2. When k is divided by 3, the remainder is 2

n=k(k+7)=k(k+1+6) --> n=k(k+1)+6k
6k is always divisible by 6. Need to show that k(k+1) is also divisible by 6.

When k is divided by 3, the remainder is 2, then k=2,5,8,11,...

for k=2,5,8,11,..., k(k+1) is always divisible by 6:
k=2: k(k+1)=2*3=6
k=5: 5*6
k=8: 8*9
k=11: 11*12

k(k+1) and 6k are both divisible by 6, therefore n=k(k+1)+6k divisible by 6. sufficient

The answer is B.
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If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 24 Oct 2017, 10:00
DanielAustin wrote:
I timed my response to work the problem--while knowing the answer--from the simplest solution I could find (quoted below) and it still took me ~3:30 mins/secs.

Any thoughts on how to cut down on the time used to solve? Has anyone else measured their response time?


KamiG wrote:
1. K is odd

It's clearly insufficient. Test k=1 and k=3.

2. When k is divided by 3, the remainder is 2

n=k(k+7)=k(k+1+6) --> n=k(k+1)+6k
6k is always divisible by 6. Need to show that k(k+1) is also divisible by 6.

When k is divided by 3, the remainder is 2, then k=2,5,8,11,...

for k=2,5,8,11,..., k(k+1) is always divisible by 6:
k=2: k(k+1)=2*3=6
k=5: 5*6
k=8: 8*9
k=11: 11*12

k(k+1) and 6k are both divisible by 6, therefore n=k(k+1)+6k divisible by 6. sufficient

The answer is B.


Hi,

If you are having a problem, solving this, you should try to master the plugin method. Actually, I solved this in a minute without writing anything.

i.e Statement 1, says k=odd, then plug in the first two odd #s i.e 1 and 3.. and you had get no and yes respectively.. not sufficient

Statement 2, qualifies statement 1 as a special type of odd ie. \(3(Q) + 2 ... (5, 8, 11)\)...if you plug in any of this numbers you will get a yes, hence statement 2 alone is sufficient...

If you actually entered consecutive odd numbers in statement 1 you would see the pattern in statement 2 already. This is a reminder that both statements are usually equivalent from some perspective and you may be able to predict the answer this way.

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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 14 Nov 2018, 15:16
TBH I didn't understand your approach Bunuel

If k is a positive integer and n = k(k + 7), is n divisible by 6?

So is n/6 = int?
(1) k is odd.
k=3
3(3+7)= 30
30/6 = int
=yes
k=1
1(8)= 8/6 = fraction
=no

Y/N are possible, so (1) --> IS

(2) When k is divided by 3, the remainder is 2.

test the first few numbers that produce a remainder of 2 when divided by 3
2, 5, 8, 11
2(2+7) = 2(9) = 18
18/6 = int

5(12)/6 = int

8(15)/6
=4(15)/3
=4(5) (int)

11(18)/6 = int

More than enough to conclude this statement is Sufficient
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 28 Feb 2019, 14:08
hi here are my two cents for this question

We need to know if n is divisible by 6, for this we need to know if the product of k(k+7) is divisible by 6

now if we have k is even then (k+7) is odd. This means that the product is divisible by 2. However if we have that k is odd then (k+7) is even. This means that the product is divisible by 2. But to confirm if its divisible by 6 we also need to know if k or k+7 is divisible by 3 also.

So Clearly statement 1 Does not help us in any way.

But With statement 2 we have that k is not divisible by 2 . or we are given that k = 3p+2, then k+1= 3p+3, this means that k+1 is divisible by 3, then we can say the other numbers that are divisible by 3 are k+4, k+7. We have earlier established that no matter what the nature of k i.e. ( Even/ odd ) nature the product is divisible by 2. and Statement 2 tells us that k+7 is divisible by 3. So we can say that n is divisible by 6.

Here is what i mean to say
if k is odd then k+7 is even , and from statement 2 we have established that k+7 is divisible by 3, so we have k+7 as a multiple of 6 hence n is divisible by 6,
now if k is even then from statement 2 we have established that k+7 is divisible by 3 and k is divisible by 2 so the product of k(k+7) is divisible by 6 hence n is divisible by 6.

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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 15 Apr 2019, 15:12
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+ 7K 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
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Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

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New post 29 Jul 2019, 03:34
Simpler way:
n = k(k+7) = k (k + 1 + 6) = k(k+1) + 6k -> we check if k (k+1) is divisible by 6
When k is divided by 3, the remainder is 2 -> k+1 is divisible by 3. While k(k+1) = product of two consecutive integers (k>0) -> k(k+1) even -> k(k+1) is divisible by 2.
One integer is both divisible by 2 and 3, it's divisible by 6 -> sufficient -> B
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Re: If k is a positive integer and n = k(k + 7k), is n divisible   [#permalink] 29 Jul 2019, 03:34

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