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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If k is an integer greater than 6, all of the following must be divisi
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18 Sep 2022, 22:05
18 Sep 2022, 22:05
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Expert Reply
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Question Stats:
50% (01:11) correct
50% (02:15) wrong based on 4 sessions
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If k is an integer greater than 6, all of the following must be divisible by 3 EXCEPT
A. k(k + 3)(k - 1)
B. 3k^3
C. (k+1)(k+5)(k+6)
D. (k+2)(k-2)(k+3)
E. k(k+1)(k+2)
A. k(k + 3)(k - 1)
B. 3k^3
C. (k+1)(k+5)(k+6)
D. (k+2)(k-2)(k+3)
E. k(k+1)(k+2)
Re: If k is an integer greater than 6, all of the following must be divisi
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21 Sep 2022, 03:42
21 Sep 2022, 03:42
1
Expert Reply
a) k(k+3)(k-1)
(k-1)*k*(k+1) are consecutives BUT instead of k+1, we have k+3 in k(k+3)(k-1) ..
and if k+1 is div by 3, k+3 will not and k and k-1 will also be not div by 3..
so this is NOT necessary to be div by 3
b)3k^3
3 is there in the term, so TRUE
c)(k+1)(k+5)(k+6)
if K+1 is div by 3, k+1+3=k+4 will also be div by 3
so we have consecutive terms by adding 3 to k+1
k+4, k+5 and k+6..
TRUE
d)(k+2)(k-2)(k+3)
add 3 to k-2. it becomes k-2+3=k+1..
again 3 consecutive terms k+1, k+2 and k+3
TRUE
e)k(k+1)(k+2)
3 consecutive terms already
TRUE
(k-1)*k*(k+1) are consecutives BUT instead of k+1, we have k+3 in k(k+3)(k-1) ..
and if k+1 is div by 3, k+3 will not and k and k-1 will also be not div by 3..
so this is NOT necessary to be div by 3
b)3k^3
3 is there in the term, so TRUE
c)(k+1)(k+5)(k+6)
if K+1 is div by 3, k+1+3=k+4 will also be div by 3
so we have consecutive terms by adding 3 to k+1
k+4, k+5 and k+6..
TRUE
d)(k+2)(k-2)(k+3)
add 3 to k-2. it becomes k-2+3=k+1..
again 3 consecutive terms k+1, k+2 and k+3
TRUE
e)k(k+1)(k+2)
3 consecutive terms already
TRUE
gmatclubot
Re: If k is an integer greater than 6, all of the following must be divisi [#permalink]
21 Sep 2022, 03:42
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