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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

Re: If x, y, z are three consecutive prime numbers
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27 Feb 2018, 23:48

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:

Only expression II is even in nature, and hence the correct answer is Option B.
_________________

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