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Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y

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Q: Is y!/x ! an integer? (1) (x + y)(x y) = 5! + 1 (2) x + y [#permalink] New post 22 Jun 2006, 20:30
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Q: Is y!/x ! an integer?

(1) (x + y)(x – y) = 5! + 1
(2) x + y = 121
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 [#permalink] New post 22 Jun 2006, 20:51
A it is
When we talk about X! and Y!, i think we should have x,y positive intergers.
(1) Because if (x+y)*(x-y)=5!+1 so x>y>=0
And y!/x! should not be an interger
(2) is insufficient
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 [#permalink] New post 22 Jun 2006, 23:00
(1) (x + y)(x – y) = 5! + 1
(2) x + y = 121
I think it's A.

(1) (x + y)(x – y) = 5! + 1
(x + y)(x – y) = 121
i.e. (x + y)(x – y) = 121 * 1
or (x + y)(x – y) = 11 * 11

Now sum & difference of two numbers can't be same.
So, x+y = 121, x-y = 1,
Solving we get y = 60, x = 61. SO y!/x ! is not integer.
Hence 1 is SUFF.

(2) x + y = 121
Clearly not SUFF.

So A.
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 [#permalink] New post 23 Jun 2006, 00:11
mendiratta_1812 wrote:
(1) (x + y)(x ? y) = 5! + 1
(2) x + y = 121
I think it's A.

(1) (x + y)(x ? y) = 5! + 1
(x + y)(x ? y) = 121
i.e. (x + y)(x ? y) = 121 * 1
or (x + y)(x ? y) = 11 * 11

Now sum & difference of two numbers can't be same.
So, x+y = 121, x-y = 1,
Solving we get y = 60, x = 61. SO y!/x ! is not integer.
Hence 1 is SUFF.

(2) x + y = 121
Clearly not SUFF.

So A.

gr8 explanation
first was going with C
now I see it is A
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 [#permalink] New post 23 Jun 2006, 01:19
the question says about factorials and it means both x and y are postive numbers.
but,its given (x+y)*(x-y)=121.
therefore,x-y>0
this implies tht x>y,
therefore,x!/y! is an integer.
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 [#permalink] New post 23 Jun 2006, 01:29
parvathaneni wrote:
the question says about factorials and it means both x and y are postive numbers.
but,its given (x+y)*(x-y)=121.
therefore,x-y>0
this implies tht x>y,
therefore,x!/y! is an integer.


actually the question is about y!/x!, not x!/y!, but I like your logic
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 [#permalink] New post 23 Jun 2006, 05:23
(1) can be read as x^2 - y^2 = 121
so if x and y are positive integers (form the stem), x must be > y and y!/x! cannot be an integer
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 [#permalink] New post 23 Jun 2006, 06:46
kook44 wrote:
(1) can be read as x^2 - y^2 = 121
so if x and y are positive integers (form the stem), x must be > y and y!/x! cannot be an integer
A


Used the same logic and got (A)

1:10
  [#permalink] 23 Jun 2006, 06:46
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