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]

22 Jun 2006, 21:30

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Q: Is y!/x ! an integer?

(1) (x + y)(x â€“ y) = 5! + 1
(2) x + y = 121

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22 Jun 2006, 21:51

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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23 Jun 2006, 00:00

(1) (x + y)(x â€“ y) = 5! + 1
(2) x + y = 121
I think it's A.

(1) (x + y)(x â€“ y) = 5! + 1
(x + y)(x â€“ y) = 121
i.e. (x + y)(x â€“ y) = 121 * 1
or (x + y)(x â€“ 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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23 Jun 2006, 01:11

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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23 Jun 2006, 02:19

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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23 Jun 2006, 02:29

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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23 Jun 2006, 06:23

(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
A

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23 Jun 2006, 07:46

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, 07:46

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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

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