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

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

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29 Mar 2016, 21:34
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Expert's post
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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16 Jun 2016, 13:09
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

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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29 Jan 2017, 15:34
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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02 Jul 2017, 09:15
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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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22 Jul 2017, 18:01
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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25 Sep 2017, 23:30
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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05 Oct 2017, 06:12
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.

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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21 Oct 2017, 08:05
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.

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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24 Oct 2017, 07:58
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

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

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24 Oct 2017, 09:00
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

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.

Best,

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If k is a positive integer and n = k(k + 7k), is n divisible   [#permalink] 24 Oct 2017, 09:00

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