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Director
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If n is a positive integer, is (n+1)!  (n1)! greater than [#permalink]
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12 Jun 2005, 18:21
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If n is a positive integer, is (n+1)!  (n1)! greater than (n+1)!/(n1)! ?
1) Square root of n is a positive integer
2) n is divisible by 2



Senior Manager
Joined: 17 May 2005
Posts: 271
Location: Auckland, New Zealand

Statement (1) does not give any useful information that i can see
(n+1)!(n1)! is greater than (n+1)!/(n1)! for any positive number except 1 (as (n1)! would be 0...i think!...thus making (n+1)!/(n1)! undefined).
If n is divisible by 2...the lowest possible value of n is two...
this is sufficient to say (n+1)!(n1)! > (n+1)!/(n1)!
thus, my answer is B



Director
Joined: 01 Feb 2003
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Location: Hyderabad

cloudz9 wrote: Statement (1) does not give any useful information that i can see
(n+1)!(n1)! is greater than (n+1)!/(n1)! for any positive number except 1 (as (n1)! would be 0...i think!...thus making (n+1)!/(n1)! undefined). If n is divisible by 2...the lowest possible value of n is two... this is sufficient to say (n+1)!(n1)! > (n+1)!/(n1)!
thus, my answer is B
the whole equation reduces to (n1)! * (n(n+1) 1) ... this needs to be compared with (n+1)!/(n1)! = n(n+1)
I) Square root of n is a positive integer
n can be 1 or 2 or any number greater than 2
for n =1 0! =1 (not zero)  LHS will be less than RHS
for n = 2, LHS < RHS
for n  3, LHS = 2*11 ; RHS = 12 => LHS > RHS
and for any number greater than 2, LHS > RHS
We need (2) to determine that n is greater than 2
(2) only this is not sufficient either as n can be equal to 2
and hence C



Senior Manager
Joined: 17 May 2005
Posts: 271
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yeah true sorry
made a stupid mistake
agree with C then



SVP
Joined: 05 Apr 2005
Posts: 1710

yah, agree with C. if n <3, second term is larger. if n>2, the first term is larger.
togather, n >2. so it is sufficient.
sparky nice problem..............



Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12

How is it C! I am coming up with E
we need to know that if N is positive, greater than 2.....
I know it is positive from both statements, but dont know if it is greater than 2?



Director
Joined: 03 Nov 2004
Posts: 853

I will go with C too
The equation reduces to (n1)![n(n+1) 1] > n(n+1)
From statement 1: n can be 1, 4, 9,...
when n is 1: (n1)![n(n+1) 1] is not > than n(n+1)
when n is 4: (n1)![n(n+1) 1] is > than n(n+1)
when n is 9: (n1)![n(n+1) 1] is > than n(n+1)
so statement 1 is not sufficient
From statement 2: n can be any even number such as 2, 4, 6, 8....
when n is 2: (n1)![n(n+1) 1] is not > than n(n+1)
when n is 4: (n1)![n(n+1) 1] is > than n(n+1)
so statement 2 is not sufficient
But both put together: n can be 4, 16, .....
As shown above the condition is satisfied for any number greater than 2
Hence C



Senior Manager
Joined: 30 May 2005
Posts: 373

The two values can be equal, or one can be greater than the other.
For n=1, the values are equal
For n=2, LHS < RHS
For n > 2 LHS < RHS
So we need to find if n=1 or n=2 or n > 2
I alone gives us not enough information. n could be 1,4,9 etc.
II alone gives us not enough information n could be 2,4,6 etc.
I&II together say that n could be 4,16,36 etc. which is sufficient to determine that LHS < RHS
So I'll go for C



Intern
Joined: 30 May 2005
Posts: 29

I will pick B.
Statement 1, it repets nothing but the original condition that n is a positive integer. It behaves differently for 1 and any higher positive integer. Hence Insufficient. Hence rule out option D as well.
Statement 2, says n is even. substraction operation on two positive integer ( Large positive integer small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number). Hence B is sufficient.
So answer is B. No need to look at C and E.



Senior Manager
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Location: India

Re: DS X [#permalink]
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14 Jun 2005, 11:52
sparky wrote: If n is a positive integer, is (n+1)!  (n1)! greater than (n+1)!/(n1)! ? 1) Square root of n is a positive integer 2) n is divisible by 2
I think it is C.
The stem reduces to (n1)! [ n^2 + n  1 ] > [ n^2 + n ] ?
We need to know that if (n1)! = 1 then it is NO , but if (n1) ! > 1 then YES.
Since the least n which is a sqaure of a number and a multiple of 2 is 4. (n1)! > 1 ,hece the aswer is YES always.
HMTG.



Intern
Joined: 30 May 2005
Posts: 29

Quote: tarungmat wrote:
substraction operation on two positive integer ( Large positive integer small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number).
3  2 is not greater than 3 /2
Clarification: I meant in question's context (with factorial ). we will never have a situation 32 and 3/2 according to statement 2.
Please let me know why not B is correct as per my logic.



SVP
Joined: 03 Jan 2005
Posts: 2233

Re: DS X [#permalink]
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15 Jun 2005, 21:51
sparky wrote: If n is a positive integer, is (n+1)!  (n1)! greater than (n+1)!/(n1)! ? 1) Square root of n is a positive integer 2) n is divisible by 2
(n+1)!  (n1)!=(n1)!*n*(n+1)(n1)!
(n+1)!/(n1)! =n*(n+1)
(n+1)!  (n1)!  (n+1)!/(n1)! = (n1)!*[n(n+1)1]n(n+1)
When n<=2, (n1)!<=1
(n1)!*[n(n+1)1]n(n+1)<=n(n+1)1n(n+1)=1<0
When n>=3, (n1)!>=2
(n1)!*[n(n+1)1]n(n+1)>=2n(n+1)1n(n+1)=n(n+1)1>0
(1) n is square of an integer, n could be 1 or 4 or greater, insufficient
(2) n is even, n could be 2 or 4 or greater, insufficient
Combined, n could only be 4 or greater, sufficient. (It's always >0 when n>=3)
(C)
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Director
Joined: 14 Feb 2005
Posts: 999
Location: New York

Quote: Square root of n is a positive integer n can be 1 or 2 or any number greater than 2
How can N be 2..N can be 1or 4 or 9
Sq root of 2 is 1.41414 some thing which is not a positive integer
correct me if i am wrong...
Cheers










