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Re: If k is a positive integer and n = k(k + 7k), is n divisible
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29 Mar 2016, 22:34
iliavko wrote: VeritasPrepKarishmaHello 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
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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
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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
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02 Jul 2017, 10:15
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
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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
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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
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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 6Not 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
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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+ 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 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
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24 Oct 2017, 08:58
I timed my response to work the problemwhile knowing the answerfrom 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
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24 Oct 2017, 10:00
DanielAustin wrote: I timed my response to work the problemwhile knowing the answerfrom 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. Best,



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Re: If k is a positive integer and n = k(k + 7k), is n divisible
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14 Nov 2018, 15:16
TBH I didn't understand your approach BunuelIf 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
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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. Probus
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Re: If k is a positive integer and n = k(k + 7k), is n divisible
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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+ 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
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Re: If k is a positive integer and n = k(k + 7k), is n divisible
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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




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