Find the value of $abc$. 
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.

In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?
 A: Let $M=396396$. Because $599\times 499<M$, we infer that $a>5$. Because $11|M$, we infer that $a+c-b=11$. Finally, $3|M$ and so $3|(a+b+c)$. So let's consider the possibilities for $a+b+c$:
$$
a+b+c=2b+11\in\{12,15,18,21,24,27\}
$$
which yields $b\in\{2,5,8\}$ which corresponds to $a+c\in\{13,16,19\}$. $19$ is too high because $c<a$. If $b=5$, then it must be that $a=9$ and $b=7$. You can verify that this doesn't work. Thus
$$
b=2\implies a+c=13\implies (a,c)\in\{(7,6),(8,5),(9,4)\}.
$$
Only $\boxed{abc=924}$ works and that's our answer.
Edit: I left out the case $a+c=b$. Let's look at it
$$
a+b+c=2b\in\{0,6,12,18\}\implies b\in\{0,3,6,9\}\implies a+c\in\{0,3,6,9\}.
$$
$0$ and $3$ are too small, and so $(a,c)\in\{(6,0),(6,3),(7,2),(8,1)\}$. Eliminate these by inspection.
A: HINT: we have $$2^2\cdot 3^2\cdot 7\cdot 11^2\cdot13$$
A: Note that $abc=100a+10b+c$ and the same decomposition for $cba$
A: Hint: 
Let $[xyz]$ denote a thre-digit number. Then you want $a,b,c$ such that $[abc]\cdot[cba]=396396$, with $a>c$.
Write $[abc]=100\cdot a+100\cdot b+1\cdot c$ and $[cba]=100\cdot c+10\cdot b+1\cdot a$, then
$$[abc]\cdot[cba]=(100\cdot a+100\cdot b+1\cdot c)\cdot(100\cdot c+10\cdot b+1\cdot a)$$
$$=10\ 000\cdot ac+1000\cdot(ab+bc)+100\cdot(a^2+b^2+c^2)+10\cdot(ab+bc)+ac$$
and this should equal $396396$. Can you continue from here?

Edit:
Listing these terms as follow:


*

*$10\ 000$: $ac$

*$1000$:  $\ \ \ ab+bc$   

*$100$: $\ \ \ \ \ a^2+b^2+c^2$

*$10$: $\ \ \ \ \ \ ab+bc$ 

*$1$: $\ \ \ \ \ \ \ \ ac$


You can see some patterns, e.g. the terms $ac$ and $ab+bc$ occur twice. Now compare this with the structure of $396396$ and consider the prime factorization of $396396$.
A: As the previous answerss said, expand $[abc][bca]$ into powers of 10's. This gives $$ac.10^4+(ab+bc).10^3+(ab+bc+ac).10^2+(ab+bc).10+ac$$
[WARNING:I am using elimination of possibilities. No constructive math here except for logic]
Now lets take a look at all possible values c and a can take. The product of a and c should give 6 in the one's digit and $a>c$. So we immediately eliminate $(4,4)$ and $(5,something)$. Since $c<a$ the only possibilities left are $(4,9), (2,3),(1,6),(7,8)$. We can now eliminate $(7,8)$ as its obvious the $10^4$ term will exceed 39. So we've got 3 possibilities for a and c. Taking $(2,3)$ or $(1,6)$ would leave too many values that b can take as the $10^4$ term is too small. however, the $(4,9)$ gives us 36 in the $10^4$ term and so the only values b can take would be 1 and 2. It will be better to eliminate this set  first. By looking at the expansion, we can also immediately guess that 1 would be too small.
Let's substitute 2. $429$x$924$ is in fact $396396$. 
I know this is guess work but from an examination point of view, this would be the optimal and simplest solution, maybe the question is structured that way. If you had more time on your hand, it won't take too long to substitute the other 2 cases using a similar logic if $(4,2,9)$ was also eliminated. 
