The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? 
The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways? 

I've seen the answer to the question, and there is only one way: 
Since $555, 555 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \cdot 7 \cdot 37) (5 \cdot 11 \cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?
Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish. 

We can write $$555,555=5\times111,111$$ and notice that $111=37\times3$, so we have $$555,555=5\times37\times3003$$ and since $1001=7\times11\times13$, the prime factorization is $$555,555=3\times5\times7\times11\times13\times37.$$ 
If we multiply each of the three combinations $11\times13$, $11\times37$ and $13\times37$ by $3$, we see that only $3\times(13\times37)$ exceeds $1000$ which is a four-digit number.
Hence $37$ must pair with two other one-digit numbers, and $11\times13$ must pair with either $3$ or $5$ since $7\times11\times13>1000$.
If $11\times13$ pairs with $3$, then the other product must be $$37\times(5\times7)>1000$$ which is not a three-digit number. 
Therefore, the only possible combination for $555,555=P_1\times P_2$ is $$P_1=5\times11\times13,\quad P_2=3\times7\times37.$$
