If a no. has 72 factors, find min/Max no. of Prime Factors possible for it.?

Answer 1

(I assume by factors you mean divisors! Because otherwise the wuestion is 'self answering": if a number has 72 factors then it has 72 factors.) If a number has a factor of p^k in it, then its divisors can have a factor of p^m where 0<=m<=k, or (m+1) possibilities. You then multiply all these possibilties over all the primes.

So if there are 72=3*3*2*2*2 divisors then the number could be a^2*b^2*c*d*e where {a,b,c,d,e} are different primes. This would yield 7 prime factors (a*a*b*b*c*d*e). That is the min. For the max, you would have a^71 with 71 prime factors and 72 divisors {1, a, a^2, a^3, ... , a^71}

Answer 2

Theoretically, the number could be the product of 72 prime factors (2 * 3 * 5 * 7...), so the maximum is 72. On the other hand, it could be a power of 2 (e.g. Factors are 2, 4, 8, 16,..., only 2 is prime) so the minimum would be 1

Ans: Min = 1, Max = 72

Answer 3

Each of the 72 factors could be a prime number so the maximum number of prime factors is 72.

If you count each of the 72 factors as a prime factor even though the same factor is repaeted 72 times as in 2^72, then the minimum is 72, but if you count a factor that is used repeatedly only once, then the minimum nuber of prime factors is 1.

Answer 4

Let g(n) be the number of factors of n.
That means if n is prime number p^m, then
g(p^m) = m + 1.

We know that n = (p1^m1)(p2^m2).....(pk^mk)

Therefore, g(n) = g(p1^m1)g(p2^m2) .... g(pk^mk)
g(n) = (m1 + 1)(m2 + 1) ..... (mk + 1)

It is given that g(n) = 72, and
72 = 2 x 2 x 2 x 3 x 3

At this point I got stuck and am not sure what to do. I apologize.