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Is product 2*x*5*y an even integer? a.) 2 + x + 5 + y is an

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Is product 2*x*5*y an even integer? a.) 2 + x + 5 + y is an [#permalink] New post 02 Aug 2008, 02:17
Is product 2*x*5*y an even integer?

a.) 2 + x + 5 + y is an even integer
b.) x - y is an odd integer

pls give explanation...
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Re: good one... [#permalink] New post 02 Aug 2008, 06:27
arjtryarjtry wrote:
Is product 2*x*5*y an even integer?

a.) 2 + x + 5 + y is an even integer
b.) x - y is an odd integer

pls give explanation...


2*x*5*y will be an even integer if x and y are integer .....

statement 1 : x+y+7 is even, x+y has to be odd so that odd + odd = even
x+y is odd, take x = 3.5 and y = 1.5 .... not suff
statement 2 : x-y is odd, take x = 3.5 and y = 0.5 ... not suff

combine m and n are positive integers.
x+y = 2m+1
x-y = 2n+1

x = m+n+1 ----> interger
y = m-n ---> integer

together suff... option C
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Re: good one... [#permalink] New post 02 Aug 2008, 10:28
arjtryarjtry wrote:
Is product 2*x*5*y an even integer?

a.) 2 + x + 5 + y is an even integer
b.) x - y is an odd integer

pls give explanation...


IMO D
Any number multiplied by 2 is even
hence irrespective of choices we can determine the answer

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Re: good one... [#permalink] New post 02 Aug 2008, 10:37
spriya wrote:
arjtryarjtry wrote:
Is product 2*x*5*y an even integer?

a.) 2 + x + 5 + y is an even integer
b.) x - y is an odd integer

pls give explanation...


IMO D
Any number multiplied by 2 is even
hence irrespective of choices we can determine the answer


OOPS silly mistake :roll: i left out possibility of fractions
just occurred to me after seeing the posts

I would re analyse this problem :
consider (1) 2+x+y+5=n n is a even integer => x+y=n-7 => x,y can be fractions or integer => insufficient
consider (2) x-y=m m is odd integer
but here also x,y can be either integer or fraction => insufficient

now consider both (1) and (2) => 2x=m+n-7=> x=(m-7+n)/2 m-7 is even and n is even hence x is integer

also 2y= n-7-m => y=(n-7-m)/2 => m+7 is even n is even hence y is integer
=> 2*x*y*5 is even integer
hence (C) is IMO. :)

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Re: good one... [#permalink] New post 02 Aug 2008, 22:37
removing the possibility of fractions was critical. this is how i did it, tell me if this sound good.

2+5+x+y = e
=> x+y = o
also given, x-y = o

adding the two, 2x must be o+o = even

if 2x is even, x must be integer. if x is integer, y is also integer, which means 10*x*y is even (ending with zero)

hence ans i C
Re: good one...   [#permalink] 02 Aug 2008, 22:37
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Is product 2*x*5*y an even integer? a.) 2 + x + 5 + y is an

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