Question

39. The number of prime facters in the product (1/6)^12(3/4)^15 8^25 is​

Answer 1

Answer:

6 - 24 6 - 25 6 - 26 6 - 27 6 - 28 6 - 29 6 - 30 6 - 31 6 - 32 6 - 33 6 - 34 6 - 35 6 ... 65% factors of a number which are prime numbers 31% only divisible by one ...

Answer 2

Let N be a composite number and a,b & c are its prime factors. Then :

N = a^p * b^q * c^r

Following hold good :

1. Number of factors = (p+1)(q+1)(r+1)
2. Number of unique factors = 3
3. Number of prime factors = p+q+r
4. Sum of factors = (a^0+a^1+..+a^p)(b^0+b^1+..+b^q)(c^0+c^1+..+c^r)
5. Product of factors = N^(Number of factors/2)

Now come to your question  (1/6)^12 * (3/4)^15 *8^25 i

it comes down as 3^3 * 2^33

so total prime factor = 36