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 \(p_1\), \(p_2\) and \(p_3 \) such that \(p_1 + p_2 + p_3 = a+b+c \)?
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
Yes, the question as it stands, is confusing but trivial
a,b,c ARE p1, p2, p3
am I missing something?