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x,y,z are positive integers

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x,y,z are positive integers

by uptowngirl92 » Wed Jul 22, 2009 5:17 am
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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by vinayakdl » Wed Jul 22, 2009 6:08 am
I will go with A

simplifying x!+x/z=y---1

z = (x/(y-x!))---2

1) as per 2 above the if x is factor of y then

z = (x/(ax-x!)) which means z = 1 since all are integers

2) is not sufficient (4,26,2),(4,28,1)

good Q, what is the source?


Vinayak
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by abhinav85 » Wed Jul 22, 2009 8:15 am
z = (x/(ax-x!)) which means z = 1 since all are integers
how u get this man???[/spoiler]
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by vinayakdl » Wed Jul 22, 2009 8:20 am
abhinav85 wrote:
z = (x/(ax-x!)) which means z = 1 since all are integers
how u get this man???[/spoiler]
Here is my reasoning.

x!+x/z=y

x/z = y-x!

x = z(y-x!)
z = x/(y-x!)

since x is a factor of y we can write y = a*x
so
z = x/(ax-x(x-1)!) = 1/(a-(x-1)!)
and since z is supposed to be an integer so are 'a' and 'x'
z will be '1'. Let me know if i am missing something.

uptowngirl92 can you please post OA?

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by uptowngirl92 » Thu Jul 23, 2009 1:18 am
Guys OA is C.Could somebody please explain?
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explanation to OA

by vyomb » Thu Jul 23, 2009 3:19 am
Can some one pls post the explanation to the OA?
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explanation to OA

by vyomb » Thu Jul 23, 2009 3:21 am
Can some one pls post the explanation to the OA?
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by vinayakdl » Thu Jul 23, 2009 12:18 pm
uptowngirl92 what is the source of this Q?

Does the answer come with any explanation?
I think OA might be wrong.

Thanks
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by mehravikas » Thu Jul 23, 2009 4:54 pm
Answer should be A.

Z has a value of 1 for x to be a factor of y
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