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Math Expert V
Joined: 02 Sep 2009
Posts: 58371
If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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Difficulty:   5% (low)

Question Stats: 86% (01:32) correct 14% (01:40) wrong based on 97 sessions

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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. 4
B. 5
C. 11
D. 15
E. 18

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examPAL Representative P
Joined: 07 Dec 2017
Posts: 1140
Re: If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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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

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.
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Senior SC Moderator V
Joined: 22 May 2016
Posts: 3545
If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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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

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.

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
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.

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Status: It's near - I can see.
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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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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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Target Test Prep Representative G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2816
Re: If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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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.

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Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4777
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
Re: If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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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

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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