There is a method to find the factors of a number as well its products.
Let us take an example for better understanding.
e.g. let us find the factors of 18
Step I:18= 2*3*3 (prime factorization)
18= (2^1) * (3^2)
Step II: Increment the powers of each prime factor by 1. i.e. power of 2 is 1, increment it by 1. Also, power of 3 is 2, increase it by 1.
Step III: Number of factors of 18 = (1+1)*(2+1) = 2 * 3 = 6
now let us try to find out the product of these factors
Step IV: product of factors of 18 = 18^ (number of factors/2)
= 18 ^ (6/2)
= 18 ^ 3
= 5832
REASON:
The factors of 18 are 1, 2, 3, 6, 9, 18.
Product of factors will be = 1*2*3*6*9*18
(we can rewrite this as..)Product of factors = (1*18)*(2*9)*(3*6) = 18*18*18 = 18^3
So....
if we want to write a generic formula, then, it can be written as
" If a positive integer 'n' has 'x' factors,
then the product of all the factors = n ^ (x/2)"SOLUTION FOR THE PROBLEM No.1
2^18* 3^12 = Product of factors of N
(2^6)*(2^12)*(3^12) = prod of factors of N
(2^12/2)*(6^12) = (6* 2^1/2)^12 = ([72]^1/2)^12 = 72 ^ (12/2)
Hence number of factors = 12
= (a+1)*(b+1)
= (2+1) * (3+1)
Hence there can be only one possible N, i.e. 72
Ans: B
SOLUTION FOR THE PROBLEM No.2
(2^9)* ( 3^9) = Product of factors of N
= 6^9
= {(36)^1/2} ^9
(this sqr root sign does not work... )
= (36)^ 9/2
hence N= 36
9 = (a+1)*(b+1)
= (2+1) * (2+1)
a= 2, b=2
Hence only possible answer, i.e.1
Ans: B