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.

  • $\begingroup$ Why not just search? There aren't that many primes to go through, after all. $\endgroup$ – lulu Jul 25 '20 at 19:38

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 $5,97$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.