# 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?

• It should be easy enough to at least approximate the number of digits in $m,n$. Also, it is clear that both $m,n$ are divisible by $9$, hence by $99$.
– lulu
Dec 22 '20 at 13:59
• Yes you can approximate, but I am looking for a solution which is elegant. I can however, guess that $m$ and $n$ will have $4$ or $5$ digits. Dec 22 '20 at 14:00
• As to the search, one knows that one of the factors is actually divisible by $99\times 37=3663$. That makes the search very fast (there are only about $20$ numbers you need to test).
– lulu
Dec 22 '20 at 14:14
• $n$ is divisible by $2$ but not $4$, so each factor has an even digit at one end and an odd one at the other; using that logic and searching the five-digit factors of the number, I found $28215$ Dec 22 '20 at 14:15
• @J.W.Tanner that is clever, I didn't think like that !! I think I got the explanation. Dec 22 '20 at 14:16

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 $$m + n = 79497$$.

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 $$m=28215, n=51282$$.