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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 \(p_1 + p_2 + p_3 = a+b+c \)?

 Nov 3, 2020
 #1
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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

 Nov 3, 2020
 #2
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Yes, the question as it stands, is confusing but trivial

a,b,c  ARE p1, p2, p3 

 

am I missing something?

 Nov 4, 2020
 #3
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I'm not too sure :(

 Nov 5, 2020

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