prime numbers

[vishal_2804]

If x, y, and z are consecutive prime numbers and x < y < z, which of the following must be true?

I. (x+y)/2 is an integer.

II. xyz has exactly 8 factors.

III. y - x = 1


a) I only

b) II only

c) III only

d) I and II only

e) II, and III

[Anju@Gurome]

If x, y, and z are consecutive prime numbers and x < y < z, which of the following must be true?

I. (x+y)/2 is an integer.
II. xyz has exactly 8 factors.
III. y - x = 1

As x, y, and z are prime, number of factors of xyz is (1 + 1)(1 + 1)(1 + 1) = 8
So, II must be true.

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

• I. (x + y)/2 = (2 + 3)/2 = 2.5 ---> Not an integer ---> I may not be true

Consider x = 3, y = 5, and z = 7

• I. y - x = (5 - 3) = 2 ---> III may not be true

The correct answer is B.

[vishal_2804]

can u plz explain the logic for As x, y, and z are prime, number of factors of xyz is (1 + 1)(1 + 1)(1 + 1) = 8.

Thanks

[Anju@Gurome]

can u plz explain the logic for As x, y, and z are prime, number of factors of xyz is (1 + 1)(1 + 1)(1 + 1) = 8.

Thanks

If the prime factorization of a positive integer N is (p^a)*(q^b)*(r^c)..., the number of different positive factors of N is given by (a + 1)(b + 1)(c + 1)...

This is because any factor of N,

• may contain either pâ�° or p¹ or p² ... or pᵃ ---> (a + 1) ways to select or not select p
  may contain either q� or q¹ or q² ... or qᵇ ---> (b + 1) ways to select or not select q
  may contain either r� or r¹ or r² ... or rᶜ ---> (c + 1) ways to select or not select r
  Ans so on...

Note that when none of the prime factors are selected (p�, q�, r�...), we get the factor 1 and when all the prime factors are selected (pᵃ, qᵇ, rᶜ ...), we get N itself as factor.

Hence, the steps for finding out the number of different positive factors of any integer N are...

• 1. Find out the prime factorization of N
  2. Increment the powers of all the prime factors by 1.
  3. Multiply the incremented powers.

For example,

• Number of different positive factors of 12 = (2²)*(3¹) is (2 + 1)*(1 + 1) = 6
  Number of different positive factors of 360 = (2³)*(3²)*(5¹) is (3 + 1)*(2 + 1)*(1 + 1) = 24

In this case prime factorization of xyz is (x¹)*(y¹)*(z¹)
So, number of different positive factors of xyz is (1 + 1)*(1 + 1)*(1 + 1) = 8

Hope that helps.

[mgm]

Thanks for the response Anju ... does this formula hold even with multiplication of prime numbers with a common factor.

E.g. 13 , 17 , 19 ... each of the primes have only 2 factors (13 & 1 / 17 & 1 / 19 &1) whereas the product should only have 4 factors and not 8 factors.. not sure what am I missing oh 13*17 is also a factor .. so that adds 2 more factors .. so I guess since we have a common factor we aren't we counting it multiple times ?

Also the optimus prime 2 was a good.. 1) holds true for every other prime also holds true for (y + z)/2..