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# If x,y,z are positive integers,and x!+x/z=y,then what is the

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Director
Joined: 25 Oct 2008
Posts: 591
Location: Kolkata,India
If x,y,z are positive integers,and x!+x/z=y,then what is the [#permalink]

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22 Jul 2009, 05:17
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If x,y,z are positive integers,and x!+x/z=y,then what is the value of z?
1. x is a factor of y
2. z<x

Guys,plugging in got me no where..Ijust went blindly plugging in random nos,didnt succeed and ended up getting the answer wrong.any ideas?
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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Senior Manager
Joined: 23 Jun 2009
Posts: 357
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: x,y,z are positive integers [#permalink]

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22 Jul 2009, 05:38
1
KUDOS
1 is suff.
y = 1.2.3....x + x/z
y= x( 1.2.3..(x-1)+1/z) so if x is a factor of y then 1/z must be integer so z=1

2 is insuff.
since x! is integer and y is integer x/z must be integer. And x>=z. This only says that x is not equal to z. We can not find the value of z.

Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil
Re: x,y,z are positive integers [#permalink]

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22 Jul 2009, 05:54
I got A as well

y = 1*2*3*4*...*x + x/z
y = x*(1/z + 1*2*3*4*...)
y/x = 1/z + 1*2*3*4 ------->> Final Expression

1) Since y/x and 1*2*3*4*... are both integers, then 1/z must be integer. The only value for z, such as 1/z is an integer is 1. Thus SUFF.

2) Using the same final expression, the fact that z<x does not help in anything.

A.
Director
Joined: 25 Oct 2008
Posts: 591
Location: Kolkata,India
Re: x,y,z are positive integers [#permalink]

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23 Jul 2009, 01:18
Guys OA is C.Could somebody please explain?
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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Senior Manager
Joined: 23 Jun 2009
Posts: 357
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: x,y,z are positive integers [#permalink]

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23 Jul 2009, 01:32
I don't think the answer is C. I think the OA is wrong.
Manager
Joined: 15 Apr 2008
Posts: 50
Location: Moscow
Re: x,y,z are positive integers [#permalink]

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23 Jul 2009, 03:36
Initially have chosen A as well, but... on the second thought it's really C:
$$x!+\frac{x}{z}=y$$
$$x*((x-1)!+\frac{1}{z})=y$$

$$x*(\frac{(x-1)!*z+1}{z})=y$$

z can be any integer, provided x=z, which makes St1 insufficient, we need St2.
Senior Manager
Joined: 23 Jun 2009
Posts: 357
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: x,y,z are positive integers [#permalink]

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23 Jul 2009, 04:18
No. z can not be any integer.

Based on the equation you found; if x = z then equation becomes (x! + 1). But x can only be the factor of y when the value of x = 1, z=1 and y=2. In other values of x; x can not be a factor of y because x! and x!+1 are coprimes (numbers that do not have any divisors other than 1). So z becomes 1 We can find z.
Manager
Joined: 15 Apr 2008
Posts: 50
Location: Moscow
Re: x,y,z are positive integers [#permalink]

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23 Jul 2009, 04:59
maliyeci wrote:
No. z can not be any integer.

Based on the equation you found; if x = z then equation becomes (x! + 1). But x can only be the factor of y when the value of x = 1, z=1 and y=2. In other values of x; x can not be a factor of y because x! and x!+1 are coprimes (numbers that do not have any divisors other than 1). So z becomes 1 We can find z.

Agree with you. Have totally missed the condition that x is a factory of y.
Re: x,y,z are positive integers   [#permalink] 23 Jul 2009, 04:59
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