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

Question

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

DavidTutorexamPAL · Accepted Answer

As we're given mostly number-property related information (even, odd, prime) we'll look for a theoretical answer.
This is a Logical approach.

Since a prime number is always odd, then x + y + z is even + odd + odd = even.
(B), (C), (D) are eliminated.
(A) 4 is too small as it can't be the sum of 3 positive integers.

(E) must be our answer.

generis · Answer

I agree with DavidTutorexamPAL that number property theory is very efficient. We have 0 < x=even < y=odd < z=ODD prime z must be odd. All primes after 2 are odd. If z = 2, y must = 1, and there is no value left for x because (x > 0) x + y + z = (E + O _1 ) + O _2 = (O _3 ) + (O _2 ) = E That said, a few real number trials will establish a pattern. x can be no smaller than 2 2 + 3 + 5 = 10 4 + 5 + 7 = 16 6 + 7 + 11 = 24 8 + 9 + 11 = 26 All are even, and all are > 4 This method is not foolproof for some variable questions (here, number property restrictions do make it foolproof). But if you're stuck, it gives you a basis from which to work. Answer E

EgmatQuantExpert · Answer

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

AkshdeepS · Answer

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)

JeffTargetTestPrep · Answer

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

Abhishek009 · Answer

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

kaziselim · Answer

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.

bumpbot · Answer

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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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