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# odds and evens

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Joined: 11 May 2008
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28 Aug 2008, 20:28
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Is product 2*x*5*y an even integer?

1. 2 + x + 5 + y is an even integer
2. x - y is an odd integer
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Joined: 07 Nov 2007
Posts: 1800
Location: New York

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28 Aug 2008, 21:03
arjtryarjtry wrote:
Is product 2*x*5*y an even integer?

1. 2 + x + 5 + y is an even integer
2. x - y is an odd integer

1)
x and y can be integers or fractions..
x=3.5 y=1.5 is not even integer
x=3 y=2 sum is even integer

insufficient

2) x - y
x and y can be integers or fractions..

combine
x+y= odd

x-y = odd

2x= odd+odd= even
2y= odd-odd= even

2*x*5*y = even*5* even/2 = integer.

So..
C.
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Last edited by x2suresh on 29 Aug 2008, 21:29, edited 1 time in total.
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Joined: 14 Aug 2007
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28 Aug 2008, 21:18
arjtryarjtry wrote:
Is product 2*x*5*y an even integer?

1. 2 + x + 5 + y is an even integer
2. x - y is an odd integer

E for me.

1) as 2 is even 1) implies that , x+y+5 = even, which implies that
x+y is odd.
3.5 + 3.5 => not even integer.
2 + 3 => even integer.

Insuff

2) 4.5 - 3.5 = 1 => not an even integer
4 + 5 => even integer

Insuff.

Combined:
again, x & y can be fractions or can be integers
so insuff

E
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Joined: 17 Jun 2008
Posts: 1548

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28 Aug 2008, 23:15
I will go with C.

If x + y is an odd integer and x - y is also an odd integere then both x and y must be integers. If x and / or y are fractions, the above conditions will not satisfy.
Current Student
Joined: 11 May 2008
Posts: 556

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28 Aug 2008, 23:21
ans iS C. good expln suresh!!!+1
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Joined: 28 Jul 2004
Posts: 136
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth

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29 Aug 2008, 20:47
answer by Suresh is correct but explanation is wrong.

if x + y = odd and x - y = odd implies that

2x = odd + odd = even = 2 * integer -> x is an integer

similarly 2y = odd - odd = even inetger = 2 * integer -> y is an integer

hence 2*x*5*y is an integer and (C) is the answer !!
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kris

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Location: New York

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29 Aug 2008, 21:29
krishan wrote:
answer by Suresh is correct but explanation is wrong.

if x + y = odd and x - y = odd implies that

2x = odd + odd = even = 2 * integer -> x is an integer

similarly 2y = odd - odd = even inetger = 2 * integer -> y is an integer

hence 2*x*5*y is an integer and (C) is the answer !!

Agreed. Thanks for pointing out.

I modified that.
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Re: odds and evens   [#permalink] 29 Aug 2008, 21:29
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