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When the digits in the number $2005$ are reversed we obtain the number $5002,$ and $5002 = a \cdot b \cdot c$, 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 $p_1 + p_2 + p_3 = a+b+c$?

My work:

So, I know that one of $p_1, p_2, p_3$ must be even, and the only even prime is $2.$ We can WLOG let $p_1=2,$ so $p_2+p_3=102$ (because $p_1+p_2+p_3=104$).

Where I'm stuck:

How do you proceed to find all the positive integers for which $p_1 + p_2 + p_3 = a+b+c$?

Any hints/solutions would be appreciated! Thanks.

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  • $\begingroup$ Why not just search? There aren't that many primes to go through, after all. $\endgroup$ – lulu Jul 25 '20 at 19:38
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Now you need to find all pairs of odd primes that add to $102$. It helps if you know the small primes by heart. The first pair is $5,97$

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  • $\begingroup$ I am getting (13,89); (19,83); (23,79); (29,73); (31,71); (41,61); (43,59). You can also have the same thing with the numbers reversed. $\endgroup$ – sshi Jul 25 '20 at 19:46
  • $\begingroup$ @sshi: those look right. As we are just interested in the number of products you should not count the reversals. $\endgroup$ – Ross Millikan Jul 25 '20 at 19:52
  • $\begingroup$ I'm not understanding the part where it says that the sum of the primes is equal to a+b+c? What is the meaning of this? $\endgroup$ – sshi Jul 25 '20 at 20:05
  • $\begingroup$ $a,b,c$ are the primes that divide $5002$. We are looking for numbers that have three prime factors that add to the same $104$ as $a,b,c$ do. As you say, one must be $2$, so the other two have to add to $102$ $\endgroup$ – Ross Millikan Jul 25 '20 at 20:54

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