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If x, y, z are three consecutive prime numbers

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If x, y, z are three consecutive prime numbers  [#permalink]

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New post Updated on: 13 Aug 2018, 01:37
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (02:03) correct 37% (01:54) wrong based on 126 sessions

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e-GMAT Question:



If x, y, z are three consecutive prime numbers where x<y<z, which of the following is even?

    i) \(x^2+y^2+z^2\)
    ii) \((x + y) * (y + z)^2 * (x + z)^3\)
    iii) \((x+3)^2 + y^2 + z\)


A) Only I
B) Only II
C) Only III
D) Both I and II
E) Both II and III

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Originally posted by EgmatQuantExpert on 27 Feb 2018, 08:36.
Last edited by EgmatQuantExpert on 13 Aug 2018, 01:37, edited 2 times in total.
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Re: If x, y, z are three consecutive prime numbers  [#permalink]

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New post 27 Feb 2018, 08:47
B) Only II, is the correct answer.

i) try to put x=2, y=3 and z=5 => E
Now try to put x=3, y=5 and z=7 => O

ii) this expression will always have one summation even so the complete expression will be even for all the values of x, y and z.

iii) repeat the process done for i, once it will be Odd and in the other it will be even.

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Re: If x, y, z are three consecutive prime numbers  [#permalink]

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New post 27 Feb 2018, 09:56
Top Contributor
EgmatQuantExpert wrote:

Question:



If x, y, z are three consecutive prime numbers where x<y<z, which of the following is even?

    i) x² + y² +z²
    ii) (x + y)(y + z)²(x + z)³
    iii) (x + 3)² + y² + z


A) Only I
B) Only II
C) Only III
D) Both I and II
E) Both II and III


Test some sets of values
Let's start with x = 2, y = 3 and z = 5
We get:
i) x² + y² + z² = 2² + 3² + 5² = 4 + 9 + 25 = 38 = EVEN
ii) (x + y)(y + z)²(x + z)³ = (2 + 3)(3 + 5)²(2 + 5)³ = (5)(8)²(7)³ = EVEN
iii) (x + 3)² + y² + z = (2 + 3)² + 3² + 5 = 25 + 9 + 5 = 39 ODD
Since iii) is not true, we can ELIMINATE C and E

Now let's test x = 3, y = 5 and z = 7
We get:
i) x² + y² + z² = 3² + 5² + 7² = 9 + 25 + 49 = 83 = ODD
Since i) is not true, we can ELIMINATE A and D

By the process of elimination, we are left with B

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Re: If x, y, z are three consecutive prime numbers  [#permalink]

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New post 27 Feb 2018, 22:48

Solution:



    • As per the question stem, x, y, and z are three consecutive prime numbers.
    • Since prime numbers can be even or odd, there can be two possibilities, depending on the values that x, y, and z can take:
      1. One of the prime numbers is even and other two are odd
        a. In this case, x has to be 2, since it is the smallest prime number among x, y and z.
        b. y and z would be 3 and 5 respectively.
      2. All the three prime numbers are odd.
        a. This means that all the prime numbers are greater than or equal to 3.
        b. For example: x = 5, y = 7, z = 11

Keeping these 2 possibilities in mind, let us analyze each of the 3 expressions:
Image
Image
Image
Only expression II is even in nature, and hence the correct answer is Option B.
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Re: If x, y, z are three consecutive prime numbers  [#permalink]

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New post 28 Feb 2018, 05:59
EgmatQuantExpert wrote:

Question:



If x, y, z are three consecutive prime numbers where x<y<z, which of the following is even?

    i) \(x^2+y^2+z^2\)
    ii) \((x + y) * (y + z)^2 * (x + z)^3\)
    iii) \((x+3)^2 + y^2 + z\)


A) Only I
B) Only II
C) Only III
D) Both I and II
E) Both II and III



firstly what all can be the cases for x<y<z
1) x is 2 so even and y and z will be odd
2) all three are odd

so y and z are ODD for sure


lets check the equations..

    i) \(x^2+y^2+z^2\)
    a) if x is even and other two odd, E+O+O=E
    b) if all three odd, O+O+O=O

    so need not be true
    ..
    ii) \((x + y) * (y + z)^2 * (x + z)^3\)
    a) if x is even and other two odd, y+z will be even..so O*E*O=E
    b) if all three odd, E*E*E=E..

    Always true
    ..
    iii) \((x+3)^2 + y^2 + z\)
    a) if x is even and other two odd, O+O+O=E
    b) if all three odd, E+O+O=O

    need not be true

only ii

B
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Re: If x, y, z are three consecutive prime numbers &nbs [#permalink] 28 Feb 2018, 05:59
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