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Divisible problem

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Divisible problem

by gmat009 » Tue Oct 07, 2008 1:00 pm
If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k – 1 is a multiple of 3.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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by namitrajiv » Tue Oct 07, 2008 1:04 pm
A it ,

n = (k-1)k(k+1)
implies product of 3 consecutive numbers , which will always be divisible by 6
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Re: Divisible problem

by gmat2010 » Tue Oct 07, 2008 1:08 pm
gmat009 wrote:If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k – 1 is a multiple of 3.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
IMO A.

in order for n to be divisible by 6, it needs prime factorization of 2 & 3. if you look at statement 1, it's 3 consecutive # that starts with number greater than 0 since it has to be positive. Therefore, for any given value of k, n is divisible by 6.

Statement B does not give us any information about n - not sufficient.

What's the OA?
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by cubicle_bound_misfit » Tue Oct 07, 2008 1:47 pm
n consecutive nos will be divisible by n!

hence A wins.
What is OA?
Cubicle Bound Misfit
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by bhumika.k.shah » Wed Jan 20, 2010 8:30 am
the OA is A
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by tanviet » Wed Jan 27, 2010 12:57 am
how do you know that product of 3 consecutives is divided by 3.2

pls, prove.
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by thephoenix » Wed Jan 27, 2010 1:37 am
duongthang wrote:how do you know that product of 3 consecutives is divided by 3.2

pls, prove.
product of 3 consecutive no. will have a no. which is multiple of 3 and a no. which is multiple of 2

n*(n+1)*(n+2)

if n is odd , then two case arises either it is div by 3 or not ; but in any case n+1 is div by 2
now when its div by 3 -------> the no. is div by 6
when it is not div by 3-------->then either n+1 or n+2 will be div by 3
and hence the product is div by 6

if n is even ---------> div by 2
either n+1 or n+2 will be div by 3
hence the product will be div by 6

one check with no's....
1*2*3
2*3*4
4*5*6
5*6*7
7*8*9

any combination u pick it will div by 6

HTH
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by ksk4uever » Sun Apr 21, 2013 3:34 am
If n is 1 den the product would be 0 in that case how it would work..0 can be divisible by something
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by Brent@GMATPrepNow » Sun Apr 21, 2013 6:18 am
gmat009 wrote:If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k-1 is a multiple of 3.
Target question: Is n divisible by 6?

Aside: For a number to be divisible by 6, it must be divisible by 2 AND by 3

Statement 1: n = k(k + 1)(k - 1)
First recognize that k-1, k and k+1 are 3 consecutive integers

There's a nice rule that says: If there are b consecutive integers, then exactly one of them is divisible by b

Since there are 3 consecutive integers, we know that one of them is divisible by 3.
We can also conclude that at least one of them is divisible by 2.
Given all of this, the product of the 3 integers (n) must be divisible by 6
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: k-1 is a multiple of 3
Since there's no information about n, there's no way to determine whether or not n is divisible by 6
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent
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by jeffrliu » Sun Nov 10, 2019 6:07 am
Brent@GMATPrepNow wrote:
gmat009 wrote:If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k-1 is a multiple of 3.
Target question: Is n divisible by 6?

Aside: For a number to be divisible by 6, it must be divisible by 2 AND by 3

Statement 1: n = k(k + 1)(k - 1)
First recognize that k-1, k and k+1 are 3 consecutive integers

There's a nice rule that says: If there are b consecutive integers, then exactly one of them is divisible by b

Since there are 3 consecutive integers, we know that one of them is divisible by 3.
We can also conclude that at least one of them is divisible by 2.
Given all of this, the product of the 3 integers (n) must be divisible by 6
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: k-1 is a multiple of 3
Since there's no information about n, there's no way to determine whether or not n is divisible by 6
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent

The best answer to this question available on the Internet.
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