SOLUTION: Find the sum of all the prime factors of 27000001. Now, this I already solved, but is there any way to better present it rather than trial division? What I did was using sum of c

Question 1048556: Find the sum of all the prime factors of 27000001.
Now, this I already solved, but is there any way to better present it rather than trial division?
What I did was using sum of cubes property to express it as 301� 89701, then factor out 301 into 7 and 43. For the number 89701, it was purely trial division.
I need your help, thanks.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
There is an algorithm to determine the sum of the prime factors of a number
:
(1) find the prime factorization of the number ( i.e. use Fermat's Factorization Method)
:
(2) form a product in which each term is the sum of all the powers
of one of the prime factors up to the exponent on that prime
factor
:
The product is the sum of the prime factors
:
Consider 9000
:
9000 = (2^3) * (3^2) * (5^3)
:
sum of factors = (1 + 2 + 4 + 8) * (1 + 3 + 9) * (1 + 5 + 25 + 125) = 30420
:
note the first term is (2^0 + 2^1 + 2^2 + 2^3), the other two terms are determined from 3^2 and 5^3 respectively
:

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