If x, y, z are three consecutive prime numbers where x

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

This is  Question 3 of  The e-GMAT Number Properties Marathon

Go to   Question 4 of the Marathon

CHIRANJIV87C · Answer

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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BrentGMATPrepNow · Answer

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

Cheers, 
Brent

EgmatQuantExpert · Answer

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: 
 https://e-gmat.com/blogs/wp-content/uploads/2018/02/15.jpg 
 https://e-gmat.com/blogs/wp-content/uploads/2018/02/21.jpg 
 https://e-gmat.com/blogs/wp-content/uploads/2018/02/3.jpg 
Only expression II is even in nature, and hence the correct answer is Option B.

chetan2u · Answer

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

jfranciscocuencag · Answer

Hello!

Shouldn't be written "must be" in order not to consider I?

Paras96 · Answer

Consider 2 cases,

x=2,y=3,z=5
x=3,y=5,z=5

On solving only II satisfies to be even in both cases
Hence B

bumpbot · Answer

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If x, y, z are three consecutive prime numbers where x < y < z

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