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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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Joined: 02 Sep 2009
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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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13 Mar 2018, 01:06
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Question Stats:

86% (01:33) correct 14% (01:40) wrong based on 100 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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Posts: 1155
Re: If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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13 Mar 2018, 01:28
4
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.

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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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13 Mar 2018, 09:45
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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If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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19 Mar 2018, 11:23
1

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

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23 Mar 2018, 11:18
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]

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05 Apr 2018, 11:11
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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Re: If x, y, and z are positive integers such that 0 < x < y < z and x is  [#permalink]

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31 Aug 2019, 07:32
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] 31 Aug 2019, 07:32
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