another q
another q A no. N is having 20 factors. What is the (i) minimum and (ii) maximum no. of prime factors of this no.?
read number of prime factors @
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
n/a
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