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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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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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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Re: x,y,z are positive integers [#permalink]
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22 Jul 2009, 05:38
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1 is suff. y = 1.2.3....x + x/z y= x( 1.2.3..(x1)+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.
So answer is A.



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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.



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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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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.



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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: \(x!+\frac{x}{z}=y\) \(x*((x1)!+\frac{1}{z})=y\)
\(x*(\frac{(x1)!*z+1}{z})=y\)
z can be any integer, provided x=z, which makes St1 insufficient, we need St2.



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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.



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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
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