Finding out the numbers under given conditions Let $M$ be a $2$ digit number $ab$, $N$ is a $3$ digit number $cde$ , and $X=M\times N$ is such that $9(X)=abcde$.The question is to find out the ratio $\frac NM$
I tried to solve it using trial and error and examined a number of cases but couldn't reach the answer so far.What I have got that $d,e$ cannot be simultaneously zero and $a+b+c+d+e$ is a multiple of $9$ I know there should be some logic behind this question instead of dwelling on trial and error.Any help shall be highly appreciated.Thanks.

Rewrite $9(X) = abcde$ as $9MN = abcde = 1000ab + cde = 1000M + N$, then divide by $M$ to get $9N = 1000 + \frac{N}{M}$ or $\frac{N}{M} = 9N - 1000$. Notice then that $\frac{N}{M}$ must be a whole number, call it $k$.
Replace $N$ with $kM$ to get $9kM = 1000 + k$. Since the left hand side is divisible by $k$, the right hand side must also be divisible by $k$, and $1000 + k \equiv 1000 \equiv 0 \pmod{k}$ implies that $k$ must be a divisor of $1000$. Since we know that $N$ is three-digit and $M$ is two-digit, the possible values of $k$ are $\{2, 4, 5, 8, 10, 20, 25, 40, 50\}$.
Notice that we also know that $1000 + k$ is divisible by $9$, due to $9kM = k + 1000$, so that leaves only one possibility, namely $k = 8$. Plugging this back in, we have $9\cdot 8 M = 72M = 1000 + 8 = 1008$, thus $M = \frac{1008}{72} = \boxed{14}$ and $N = kM = 8\cdot 14 = \boxed{112}$, and this is the only solution.
Indeed, $9(X) = 9(14\cdot 112) = 14112$.
