Answer :
By using arithmetic progression, it is obtained that the number of natural numbers less than 100 whose product of the number's distinct prime factors equal to 6 = 13
What is Arithmetic Progression?
A sequence of numbers is said to be in Arithmetic progression if the difference between the consecutive terms of the sequence are same.
If the first term is a and the common difference is d, then the [tex]n^{th}[/tex] term of the sequence is given by the formula
[tex]a_n = a + (n - 1)d[/tex] , where [tex]a_n[/tex] is the [tex]n^{th}[/tex] term
The natural numbers whose product of the number's distinct prime factors equal to 6 must have only 2 and 3 as distinct prime factors
The natural numbers less than 100 which have only 2 and 3 as a prime factor are multiples of 6 except 30, 60, 90
Now to calculate number of multiples of 6 less than 100, Arithmetic Progression is required
First term = 6, last term = 96, common difference = 6
[tex]96 = 6 + (n - 1) 6\\\\n - 1 = \frac{96 - 6}{6}\\\\n - 1 = 15\\\\n = 15 + 1\\\\n = 16[/tex]
Now, 30, 60 and 90 are discarded as they have 5 as a prime factor also
Number of natural numbers less than 100 whose product of the number's distinct prime factors equal to 6 = 16 - 3 = 13
To learn more about Arithmetic Progression, refer to the link-
https://brainly.com/question/6561461
#SPJ4
