Consider the 4th problem of the Project Euler:
A palindromic number reads the same both ways. The largest palindrome made from the product of two $2$-digit numbers is $9009$ = $91 \times 99$.
Find the largest palindrome made from the product of two $3$-digit numbers.
I found a solution but there are better solutions. The one given by the Project Euler community uses this fact:
Consider the digits of $P$ let them be $x$, $y$ and $z$. $P$ must be at least $6$ digits long since the palindrome $111111 = 143\times 777$ – the product of two $3$-digit integers.
My question is why is it not considering palindrome numbers with $5$ digits from $11111$ to $99999$ since they can be possibly obtained by the product of 2 3-digits numbers? I mean if it were not for the example given in the solution $111111 = 143\times 777$ we would not know if such a a palindrome exists, and if it is a 5 or 6 digits palindrome, would we?
The question asks to find the largest palindrome number made from the product of two $3$ digit numbers .
If you know that $111111=143\times 777$ (which is a palindrome and the product of $2$ three digit numbers) then you also know that the numbers you are trying to find has $\ge 6$ digits because a number with less than $6$ digits cant be the largest palindrome because $111111>$any number with less than $6$ digits .