Editorial doubt .....!

#1

“The product of all the divisors of a number N is N^x where x is the number of divisors of N.”

Now N=12, x=6 then 12^6=2985984
but product of divisor=1x2x3x4x6x12…They are not the same

#2

it is N^(x/2) where x is the number of factors
To find product of all the factors we follow the following steps
Step1: Prime factorisation N= a^pb^qc^r
Step2: Number of factors (say X)
Step3: Product of all the factors is given by = N^(X/2)
Example 2: Find the product of all the factors of 120 ?
Solution:
Step 1: Prime factorisation : = 2^3X5^1X3^1
Step2: Number of factors are ( 3+1)(1+1)(1+1)= 16
Step 3: Product of all the factors =( 23X51X31)16/2 =(120)^8

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#3

But product of all factors will become very large in order of 10^17~18. So if it is possible to further find its prime factorization and get the total number of divisors or if there is some direct method for this?

#4

this question is based on only combinatorics, there is nothing as hard about it…