SOLUTION: The product of four prime numbers (they don't necessarily need to be all different) is the same as ten times the sum of these prime numbers. What are all the possibilities for tho

Question 909060: The product of four prime numbers (they don't necessarily need to be all different) is the same as ten times the sum of these prime numbers.
What are all the possibilities for those four prime numbers?
The only possiblity that I've found so far is this:
2+3+5+5 = 15 and
2*3*5*5 = 150
What are other possibilities? How can I find out? Is there a formula?
In case, there are no more possibilities, what is the proof?
Thank you very much, all help is greatly appreciated!
P.S. Sorry for my English, I hope the question is understandable.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
look at this as a prime factorization problem
let x be an integer with four prime factors, then we know that
x = a^p * b^q * c^r * d^s where a,b,c,d are prime factors and p,q,r,s are their powers, then
The number of factors of x will be expressed by the formula (p+1)(q+1(r+1)(s+1). NOTE: this will include 1 and x itself.
We are told that x=abcd, where a, b, c, and d are four distinct prime numbers. According to the above the # of factors of x including 1 and x is (1+1)(1+1)(1+1)(1+1)=16. Excluding 1 and x the # of factors is 16-2=14.
This exercise assumes that each prime factor is distinct, this gives you an upper bound on the number of prime factors of x.

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