# Primes, Digits, and Number Theory in Competition Math

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.

• Why not just search? There aren't that many primes to go through, after all. – 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 $$5,97$$
• $a,b,c$ 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 $102$ – Ross Millikan Jul 25 '20 at 20:54