CAT4MBA

20 = 1 x 20

     = 2 x 10

     = 4 x 5

for 1 x 20 , number of prime factors 0, 21

for 2 x 10 , number of prime factors 1, 9

for 4 x 5, number of prime factors 3, 4

so minimum is 7 and maximum is 21

can u please explain ur solution.

read number of prime factors @

http://www.cat4mba.com/math-e-book/number-system

hi

the answer is - min-  1 (that i have understood that any natural no. of hte format p^19 will have 20 factors, whre p is a prime no.

the max given is - 3.

how?

rgds

Question: A number 'N' is having 20 factors. What is the (i) minimum and (ii) maximum number of prime factors of this number?

Solution:

We know that any number N can be expressed in the form:

N = (p^a) * (q^b) * (r^c) * ............., where p, q, r are distinct prime numbers.

In that case, the number of divisors of the number is given by the product:(a + 1)(b + 1)(c + 1) ............. and so on.

In the given problem, it is mentioned that the number of divisors of N is 20.

Now, the total number of ways in which 20 can be expressed as a product of one or more than one prime factors are:

20 = 20 x 1, which shows that the number should be of the form (p^19), where p is a prime number.

20 = 4 x 5, which shows that the number should be of the form (p^3) * (q^4), where p and q are distinct prime numbers.

20 = 2 x 10, which shows that the number should be of the form (p) * (q^9), where p and q are distinct prime numbers.

20 = 2 x 2 x 5, which shows that the number should be of the form (p) * (q) * (r^4), where p, q and r are distinct prime numbers.

Hence we see that the number can have a:

(i) Minimum of ONE prime factor.

(ii) Maximum of THREE prime factors.

Thank You.

Ravi Raja