very hard DS, help, help

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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Thu Oct 29, 2009 5:40 am

this is from 15 testset.

if n and k are positive integers, is n divisible by 6

a, n=k(k+1)(k-1)
b,k-1 is a multiple of 3

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Source: Beat The GMAT — Data Sufficiency | Thu Oct 29, 2009 5:40 am

mp2437: Master | Next Rank: 500 Posts; Posts: 177; Joined: Thu Aug 14, 2008 11:59 am; Thanked: 25 times

by mp2437 » Thu Oct 29, 2009 6:50 am

Statement 1: Plug in a few numbers, you can see that for any positive integer k, n is divisible by 6. Sufficient.

Or, for a more formal definition, notice that k(k+1)(k-1) are three consecutive integers, and there is always a number in the sequence that has a multiple of 2, and a subsequent number that is a multiple of 3 (i.e 1,2,3 or 2,3,4 or 7,8,9), so having a multiple of 3 and another multiple of 2 ensures that any number k will be divisible by 6.

Statement 2: No relationship to n. Insufficient.

Answer: A

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Gmatter2.0: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Fri Oct 30, 2009 10:45 pm; Thanked: 1 times; GMAT Score:580

by Gmatter2.0 » Sat Oct 31, 2009 11:24 pm

Why not the answer be D.

if n and k are positive integers, is n divisible by 6

a, n=k(k+1)(k-1)
Yes for all values of K n is divisible by 6
This is true even if K=1 because 0/6=0
i.e 0/any number is 0 , 0 is not divisible only by 0.
Sufficient:

b,k-1 is a multiple of 3

If (k-1) is a multiple of 3 then K can either be even or Odd
Case 1:
If K is even and (k-1) is multiple of 3
K(k+1)(k-1)
even (multiple of 2) * odd * Odd and multiple of 3/ --6 Yes!
Case 2:
If K is odd and (k-1) is multiple of 3 , K cannot be Zero
K(k+1)(k-1)
odd* even (a multiple of 2) * Odd and multiple of 3/ --6 Yes!
Sufficient:

So the answer is D.

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NikolayZ: Master | Next Rank: 500 Posts; Posts: 124; Joined: Thu Jun 18, 2009 5:33 am; Thanked: 35 times

by NikolayZ » Sun Nov 01, 2009 5:18 am

A, to my mind.

1) From 1 we know the relationship between n and k.
n is product of 3 consecutive integers. And at least one of them MUST be even (i.e. a multiple of 2) So, the product is divisible by 2 and 3. Hence N is divisible by 6. Sufficient.

2) Knowing that (k-1) is a multiple of 3, we cannot decide whether n is divisible by 6. Insufficient then.

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2010gmat: Master | Next Rank: 500 Posts; Posts: 371; Joined: Fri Mar 06, 2009 2:48 am; Thanked: 27 times; GMAT Score:740

by 2010gmat » Mon Nov 09, 2009 10:00 am

for me too A...whenever we have product of three consecutive integers ...the product is divisible by 6...

-1,0,1
0,1,2
1,2,3

take any combo u want...and the product would be div by 6

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palvarez: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Oct 24, 2009 4:43 pm; Thanked: 22 times; GMAT Score:710

by palvarez » Mon Nov 09, 2009 1:20 pm

It is not that hard.

Just note down this concept:

Product of any n consecutive integers is divisible by n!

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Nov 09, 2009 3:45 pm

Gmatter2.0 wrote:Why not the answer be D.

if n and k are positive integers, is n divisible by 6

a, n=k(k+1)(k-1)
Yes for all values of K n is divisible by 6
This is true even if K=1 because 0/6=0
i.e 0/any number is 0 , 0 is not divisible only by 0.
Sufficient:

b,k-1 is a multiple of 3

If (k-1) is a multiple of 3 then K can either be even or Odd
Case 1:
If K is even and (k-1) is multiple of 3
K(k+1)(k-1)

This is the data sufficiency police: STOP RIGHT THERE AND PUT YOUR MARKER UP!

How is K(k+1)(k-1) relevant to statement (2), taken alone?

You've committed one of the most serious (and common) DS crimes - you're pretending to evaluate statement (2) by itself, but you're stealing information from statement (1).

Remember step 2 of the Kaplan Method for Data Sufficiency: Consider each statement by itself (in conjunction with the stem).

The only way you could make (2) sufficient is by using the information provided in (1), which clearly indicates that (2) isn't sufficient alone.

Looking at statement (2) in a vaccuum, we have information about k, but absolutely no information about n; accordingly, there's no way it can be sufficient to answer a question about n.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
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Gmatter2.0: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Fri Oct 30, 2009 10:45 pm; Thanked: 1 times; GMAT Score:580

by Gmatter2.0 » Thu Dec 10, 2009 10:31 pm

Thank's for pointing my error, My big mistake...

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