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Re: Is XYZ an odd number [#permalink]
08 Apr 2010, 11:41
B ? With the second you can know that x must be even so any other two numbers x an even number = even number With the first one you have two option I think: if the primer number is 2 or any other
Re: Is XYZ an odd number [#permalink]
08 Apr 2010, 11:58
IMO E,
2nd statement just give information about X, we need to know primarily the Z. thus not sufficient.
now in statement 1 Any prime number - Z = Positive Odd number
as prime number - any number can be equal to odd number only when:
case 1-- prime number is 2...then 2-z = +ve odd number only when z=1 case 2-- prime number other than 2.....say 7 -z = odd then z must be even.... thus this is also not sufficient.
even if you combine no success..thus E _________________
Re: Is XYZ an odd number [#permalink]
08 Apr 2010, 14:31
Hi gurpreetsingh, can you please explain more why 2) is INSUF
Quote:
Is XYZ an odd number? Where X, Y, Z are positive integers Such that XYZ is not divisible by 4 and X>Y>Z
1) Any prime number - Z = Positive Odd number 2) X +1 = Prime
Since X is the biggest numbers, x can be 2, in this case 2>1>0 and 2x1x0 = 0 = even or if x is other number for example 10 10+1 = 11 prime number xyz = 10yz = even*number = even
but x cannot be 1 because if x= 1 y must be 0 and z = -1 since x>y>z and all must be positive integers so I think that xyz should be even
any ideas? something wrong in my way of thinking? thanks!
Re: Is XYZ an odd number [#permalink]
08 Apr 2010, 19:16
I am considering XYZ as X*Y*Z.
From the question stem, X,Y,Z are positive integers and X>Y>Z. Inference: XYZ is not divisible by 4 => XYZ do not have two even numbers, either it should be 1 even and two odd numbers or three odd numbers. Statement 1: Any prime - Z = positive odd number, So it shows Z is an even number. ( we can exclude 2 as a prime number in this situation because,the statements says it is any prime, all other primes are odd except 2)
From inference, Because Z is even, the only other possibility for XYZ is to have two odd numbers X, Y. odd*odd*even = even Sufficient
Statement 2: X +1 = Prime Because X>Y>Z and all are +ve integers, X should be greater than 1. X should be an even number to make X+1 as prime number (odd).
From inference, because X is even the only other possibility for XYZ is to have two odd numbers Y,Z. even*odd*odd = even Sufficient
Re: Is XYZ an odd number [#permalink]
09 Apr 2010, 05:50
Hi,
I am also considering XYZ as X*Y*Z.
From the question stem: 1) X,Y,Z are positive integers so 1,2,3 .... 1000000 2) X>Y>Z so X>Y>Z>0 3) XYZ is not divisible by 4, so XYZ is not a multiple of 4 or XYZ do not have two even numbers (two 2)
The real question is whether XYZ contain a 2 or is even ?
Statement 2: 2) X + 1 = Prime Prime: 2,3,5,7,11,... X could be 1,2,6,10 but as we know that X>Y>Z>0 then X can not be equal to 1, then X is even Sufficient
1) Any prime number - Z = Positive Odd number Prime Odd number: 3,5,7,11,... Prime number: 2,3,5,7,11,...
Z is positive from question stem. If prime number = 2 -> 2-Z = positive odd number then Z = 1 If prime number = 3,5,7,11 -> 3 - Z = positive odd number then Z = 2 or a multiple of 2 Insufficient
Re: Is XYZ an odd number [#permalink]
09 Apr 2010, 06:17
Sorry guys, I think I was wrong in exclding 2 for statement 1. If we consider 2 also then for the first statement will give us two answers and so it is Insufficient.
Answer should be B.
gmatclubot
Re: Is XYZ an odd number
[#permalink]
09 Apr 2010, 06:17
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