9
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ 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 $99$. $\endgroup$
    – lulu
    Dec 22, 2020 at 13:59
  • $\begingroup$ 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. $\endgroup$
    – Anonymous
    Dec 22, 2020 at 14:00
  • 1
    $\begingroup$ 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). $\endgroup$
    – lulu
    Dec 22, 2020 at 14:14
  • 1
    $\begingroup$ $n$ 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 $28215$ $\endgroup$ Dec 22, 2020 at 14:15
  • 1
    $\begingroup$ @J.W.Tanner that is clever, I didn't think like that !! I think I got the explanation. $\endgroup$
    – Anonymous
    Dec 22, 2020 at 14:16

2 Answers 2

6
$\begingroup$

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 $m + n = 79497$.

$\endgroup$
6
$\begingroup$

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 $m=28215, n=51282$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .