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If x,y, and z are positive integers such that 0 < x <

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Director
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If x,y, and z are positive integers such that 0 < x < [#permalink] New post 27 May 2005, 15:02
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A
B
C
D
E

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If x,y, and z are positive integers such that 0 < x < y < z and x is even, y is odd, and z is prime, which of the following is a possible value of x + y + z ?

A) 6
B) 11
C)19
D) 26
E) 29
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 [#permalink] New post 27 May 2005, 17:36
2+5+19 =26

D.

Just started plugging away. Probably took me 2:30 - 3:00 min. Is there an easier way to do this?
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 [#permalink] New post 27 May 2005, 17:45
tkirk32 wrote:
2+5+19 =26

D.

Just started plugging away. Probably took me 2:30 - 3:00 min. Is there an easier way to do this?


:) of course, with the right approach, it would take 30 sec, maybe less, including reading the question.
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Re: PS, 700+ level [#permalink] New post 27 May 2005, 17:56
sparky wrote:
If x,y, and z are positive integers such that 0 < x < y < z and x is even, y is odd, and z is prime, which of the following is a possible value of x + y + z ?

A) 6
B) 11
C)19
D) 26
E) 29


to minimize the time, use POE first. so eliminate B, C and E because x + y + z cannot be odd integer.

then, plug in. using plugging in, you can easily eliminate A because 6 is smaller than the required integer. so D (26) is the only possible integer.
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 [#permalink] New post 27 May 2005, 18:06
see, I am just not good enough to see that 2 odds plus an even must be even.

I know the concept, just cant seem to apply it when I need to.
Director
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 [#permalink] New post 27 May 2005, 19:10
let the

even number x =2*k
odd number = 2*k+1

prime number is odd always uneless it is 2.

add them 4*k + 1 + odd number(prime number) = some even number

x has to be atleast 2 ......answer is 26
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Re: PS, 700+ level [#permalink] New post 28 May 2005, 02:35
sparky wrote:
If x,y, and z are positive integers such that 0 < x < y < z and x is even, y is odd, and z is prime, which of the following is a possible value of x + y + z ?

A) 6
B) 11
C)19
D) 26
E) 29


Sorry guys, saw this thread late, but thought would chip in with my thoughts. My 2 cents on this one -

Min x = 2. Min y = 3. Min z = 2. Thus x + y + z >= 7. Thus A is ruled out.

You can start a process of elimination by considering the primes. Basic requirement is this:

Subtract a prime from the given option, and your answer should be odd.
If this hits, then the prime should be greatest of the three numbers.

Since odd number - prime would always be even (unless the prime is 2, which is smallest number, so inadmissible) - since all other primes are odd, no odd number can be the answer.

Thus 26.


Btw Sparky, POE is Process of Elimination (unless I am grossly mistaken :) )
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 [#permalink] New post 28 May 2005, 10:01
D it is...
sorry late on the board...

someone told me on GMAT whenever you have which of following type question start with the answer choice E and start plugging in numbers....

I did that, took about 1 min to figure D it is...
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 [#permalink] New post 29 May 2005, 09:31
took me 2-3 mins but it cleared quite a few concepts.

I think plugging in numbers might be quick if you get it quickly but it is possible that it might take time too.

x,y are odd and even.

x + y is odd.
z is prime and not =2. z is odd.

( All prime numbers are odd except 2. Here x,y are smaller than z so z cannot be 2)

Now we know that ODD + ODD = EVEN

The only even answers are 6 and 26.

6 can be easily verified as not sufficient.

Ans is 26.

HTH
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 [#permalink] New post 02 Jun 2005, 02:21
I saw this question very late. However here is my approach and answer.

0 < x < y < z

X is even, Y is odd and Z is prime. 2 is the only even prime number, and z cannot be equal to 2, and so z should be an odd prime number.

even + odd + odd = even

From the answer choices there are only 2 choices 6 and 26 that are even. If x = 2 then the answer choice cannot be 6, hence 26 is the only option.

Thanks
Sreeni
  [#permalink] 02 Jun 2005, 02:21
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