DS#170 in the GMAT Review book wrong?

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triciatorres: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Thu Dec 08, 2011 8:56 am

by triciatorres » Thu Dec 29, 2011 10:26 am

Please explain the following problem below

170. If n is a postive integer, is (n^3)-n divisible by 4?

(1) n=2k+1, where k is an integer
(2) (n^2)+n is divisible by 6

The official answer claims that A is sufficient, however can someone explain to me why?

If in (1) k=0, then n = 2(0)+1 = 1. Thus (n^3)-n = (1^3)-1 = 0, which is not divisible by 4. On the other hand if k=1, then n= 2(1)+1 = 3. Thus (n^3)-n = (3^3)-3 = 27-3= 24, which is divisible by 4 (24/4=26).

Since there are differing outcomes to Statement(1), I answered (1) as being INSUFFICIENT. Which is incorrect in the answer key. I read the answer explanation and it does not mention anything about when k=0.

I am running on the assumtion that 0 is an integer and that 0 is not divisible by 4. Are they assumptions incorrect?

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Source: Beat The GMAT — Data Sufficiency | Thu Dec 29, 2011 10:26 am

rijul007: Legendary Member; Posts: 588; Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members; Location: New Delhi, India; Thanked: 130 times; GMAT Score:720

by rijul007 » Thu Dec 29, 2011 11:17 am

triciatorres wrote:Please explain the following problem below

170. If n is a postive integer, is (n^3)-n divisible by 4?

(1) n=2k+1, where k is an integer
(2) (n^2)+n is divisible by 6

The official answer claims that A is sufficient, however can someone explain to me why?

If in (1) k=0, then n = 2(0)+1 = 1. Thus (n^3)-n = (1^3)-1 = 0, which is not divisible by 4.

I am running on the assumtion that 0 is an integer and that 0 is not divisible by 4. Are they assumptions incorrect?

0 is divisible by 4.

When you say N is divisible by M, what do you mean?
It means that when you divide N by M, you dont get any remainder

Similarly,
if you divide 0 by 4, you dont get any remainder

So 0 is divisible by 4.

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Thu Dec 29, 2011 11:37 am

n>0 and Is n(n^2-1) divisible by 4?
st(1) implies (n-1) is even as (n-1)=2k; if (n-1) is even then n is odd.
given in our question stem n(n^2-1) can be rewritten as n(n-1)(n+1) which turns into (odd*even*even)

the product of any two even numbers is divisible by 4 (because even number contains 2 as coefficient and 2*2 coefficients mean clear divisibility by 4)
Hence st(1) is Sufficient

st(2) implies n(n+1) is even. Thus, either n is even or n is odd. Not clear and Not Sufficient.

a

triciatorres wrote:Please explain the following problem below

170. If n is a postive integer, is (n^3)-n divisible by 4?

(1) n=2k+1, where k is an integer
(2) (n^2)+n is divisible by 6

The official answer claims that A is sufficient, however can someone explain to me why?

If in (1) k=0, then n = 2(0)+1 = 1. Thus (n^3)-n = (1^3)-1 = 0, which is not divisible by 4. On the other hand if k=1, then n= 2(1)+1 = 3. Thus (n^3)-n = (3^3)-3 = 27-3= 24, which is divisible by 4 (24/4=26).

Since there are differing outcomes to Statement(1), I answered (1) as being INSUFFICIENT. Which is incorrect in the answer key. I read the answer explanation and it does not mention anything about when k=0.

I am running on the assumtion that 0 is an integer and that 0 is not divisible by 4. Are they assumptions incorrect?

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LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Thu Dec 29, 2011 12:02 pm

triciatorres wrote:Please explain the following problem below

170. If n is a postive integer, is (n^3)-n divisible by 4?

(1) n=2k+1, where k is an integer
(2) (n^2)+n is divisible by 6

q.- ((n^3)-n)/4 ?

((n^3)-n))/4=(n(n^2)-1))/4=n*(n-1)(n+1)/4 ?

1) n=2k+1 then (2k+1)*(2k+1-1)*(2k+1+1)=4k(2k+1)(k+1) of course 4K(2k+1)(k+1) is divisible by 4
hence ,suff

2) (n^2)+n is divisible by 6
it means,that n(n+1)/3*2 it is divisible by 2 , but it does not mean, that it is also divisible by 4

hence, insuff

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