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Re: Is XYZ an odd number? Where X, Y, Z are positive integers [#permalink]

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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? Where X, Y, Z are positive integers [#permalink]

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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? Where X, Y, Z are positive integers [#permalink]

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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? Where X, Y, Z are positive integers [#permalink]

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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? Where X, Y, Z are positive integers [#permalink]

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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? Where X, Y, Z are positive integers [#permalink]

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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.

Re: Is XYZ an odd number? Where X, Y, Z are positive integers [#permalink]

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21 Nov 2016, 02:22

Hi Everyone. I just Formatted the Question to make it slightly more pleasing to the Eye

Heres my solution here we are given that x,y,z are all >0 and integers with x>y>z We need to check if xyz is odd or not

Statement 1 Here using test cases Let z=1 x=10 y=9 hence xyz is even let z=1 x=9 y=7 hence xyz sill be odd Clearly not sufficient

Statement 2 Here the least value of x is 3 as x>y>z and all are positive integers hence x+1=> prime number >4 Since all the prime number greater than 4 are odd hence x+1=> odd so x=> odd-1=> even Hence xyz will always be even