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

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

We know:

• x is an even number greater than 0.
o Thus x>=2.

• y is an odd number.
• Since z is a prime number greater than x, thus z is an odd number.

Now, we know the nature of x, y, and z, we can easily find the nature of (x + y+ z).

• x + y + z= Even + odd + odd
• x + y + z = Even + Even
• x + y + z = Even
The two even number in the given options are 4 and 18.

The minimum value of x and z are 2 and 3 respectively. And the sum of the minimum values of x and z, that is 2 and 3, is already greater than 4.
Hence, the value of x +y + z cannot be 4.

Therefore, the possible value of (x + y + z) is 18.

Answer: E
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
Bunuel 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. 4
B. 5
C. 11
D. 15
E. 18

Here logic wins;

Given : 0 < x < y < z and x is even, y is odd, and z is prime

x = even, that means we don't require digit "1" here.

Start with "2" as "x" is even, then next two numbers (y and z) are greater than "2".

This means the sum of x + y + z will always be greater than "9".

Consider 2 + 3 + 4 (without restriction in a sequence), so A and B are out.

Now with given considerations in the question:

x = even, y = odd, and z = prime (this prime will always be odd as we have consumed the only even prime number "2" as "x".

Therefore, even + odd + odd = even.

Our answer should be even so C and D are also out.

(E)
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
1
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Bunuel 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. 4
B. 5
C. 11
D. 15
E. 18

If x is even, y is odd, and z is prime, then the values could be:

x = 2

y = 3

z = 13

So x + y + z = 18.

Answer: E
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
Bunuel 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. 4
B. 5
C. 11
D. 15
E. 18

x + y + z =Even + Odd + Prime (Odd) = Even

Now the prime number can have 2 possibilities -

1. Odd
2. Even - Not possible as it must be > 2

Thus, the answer must be even, left with option (A) and (E)

(A) can be negated as the least value of x = 2 and considering the other values > 2 , the answer can never be 4, Hence Answer must be (E) 18
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
Bunuel 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. 4
B. 5
C. 11
D. 15
E. 18

Minimum value of x must be 2 as it is stated in the question that x is even and it is greater than 0.

y is odd

z is prime..........all the prime is odd except 2 , the lowest prime number. Since z is greater than x ..........z must be odd

even + odd + odd = even

Only option E contains an Even value.

thus the best answer is E.
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is [#permalink]
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