If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$. 
If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.

What I Tried: I found $1446921630 = 2 * 3^5 * 5 * 7 * 11^2 * 19 * 37$ , but that did not really give me useful information. 
A little information I got is that $m$ and $n$ each will have a factor of $11$ , and at-least $1$ factor of $3$, as both will be divisible by $3$ and $11$ .
I could have assumed the numbers to be of the form $10x + y$ or something, but I can't as I don't know how many digits both $m$ and $n$ will have, and that will more like Trial and Error.
Another thing is that among $m$ and $n$, one will be even and one will be odd. The odd number will start with an even digit which is not $0$, and the even number cannot end with $0$. Also $5$ should divide the odd number and it will end with $5$. That is all I could conclude.
Can anyone help me?
 A: Observe that $mn \approx 1.44 \times 10^9$, $5 \mid mn$ and $4 \not\mid mn$, so one of the two numbers is odd ending with the digit $5$ and another number is even but not divisible by $4$.  WLOG, assume $5 \mid m$ and $2 \mid n$.  Make an estimate $n \approx 5 \times 10^4$, so that $m \approx 2.88 \times 10^4$.  This drives us to the guess that the first two digits of $m$ are $2$ and $8$.  We now have $28 ?? 5$.
Use the criteria that $9$ and $11$ divides both $m$ and $n$ to exhaust all possible candidates.  We focus on these two criteria first since these criteria are invariant upon reversion of digits.  By trial, it's easy to find that $m = 28215$, so $n = 51282$, and thus $m + n = 79497$.
A: Just another approach...
$1$st, estimate the number of digits in $m,n$:
$$
\sqrt{1446921630} \approx 38038,
$$
so we expect that $m,n$ are of the form
$$
m = 10^4 a+10^3b+10^2c+10d+e, \\
n = 10^4 e+10^3d+10^2c+10b+a, \\
$$
where $a,b,c,d,e$ are just digits ($a,e \ne 0$).
$2$nd,
$$
m n = 10^8ae+10(\ldots) + ae,
$$
so:
 a) $ae\le 14$;
 b) $ae$ has last digit $0$.
From here we have only one pair for $a\le e$:
$$a=2, \;e=5.$$
$3$rd,
now $mn$ has the form
$$
mn = 10\cdot 10^8 + 10^7(2d+5b) + 10^2(...) + 10(2d+5b) + 2\cdot  5.
$$
and we have estimation:
$$
mn > 10\cdot 10^8+10^7(2d+5b).
$$
$$
mn < (2\cdot 10^4 + 10^3(b+1))(5\cdot 10^4+10^3(d+1)) \le 10\cdot 10^8 + 10^7(2d+5b)+10^6\cdot 100.
$$
from here we conclude:
 a) $34.69 <2d+5b < 44.69$
 b) last  digit of $2d+5b$ is $2$.
Definitely $2d+5b = 42$.
Therefore, only $2$ possible cases work for $b,d$:
 $b=6, d=6$;
 $b=8, d=1$.
$4$th: so we have $2$ candidates:
a) $m = 26?65$, $n = 56?62$,
b) $m = 28?15$, $n = 51?82$.
case a) is too big: even if $?=0$, we'll have $mn>14.6\times 10^8$.
case b) leads to $m=28215, n=51282$.
