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If x,y, and z are positive integers such that 0 < x < [#permalink]

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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 ?

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.

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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