+479 When the digits in the number 2005 are reversed we obtain the number 5002 and 5002=a*b*c such that a, b and c are three distinct primes. How many other positive integers are the products of exactly three distinct primes p1, p2 and p3 such that p1+p2+p3=a+b+c?
What is the ceiling of the numbers you want to check? Because there are an infinite number of integers which will have 3 distinct prime factors, such as: 9971 x 123457 x 1000003 =1,230,993,439,969,241. You can simply reverse the digits:1429699343990321
