$35 m$ and $n$ are two positive integers of reverse order (for example 123 and 321 ) such that $m n=1446921630$. Find the value of $m+n$.
35 Answer: (79497)
Clearly, $m$ and $n$ are both 5 -digit numbers.
Next, it would be helpful that we know $m n=2 \times 3^{5} \times 5 \times 7 \times 11^{2} \times 19 \times 37$.
Now since the last digit of $m n$ is 0 , we may assume $5 \mid m$ and $2 \mid n$. But the first digit of $m n$ is $1 \Rightarrow$ Last digit of $m$ is 5 (not 0 ) and last digit of $n$ is 2 (not 4,6 or 8 ).
Also, $3^{5} \mid m n$, so 9 divides at least one of $m$ and $n$. On the other hand, $9|m \Rightarrow 9| n$. Similarly $11|m \Rightarrow 11| n$.
Set $n=198 k$. Then the last digit of $k$ is 4 or 9 .
Since the remaining factors $3,7,19,37$ are odd, the last digit of $k$ must be 9 . We have only the following combinations: $k=7 \times 37$ or $3 \times 7 \times 19$ or $3 \times 19 \times 37$.
Recall that the first digit of $n$ is 5 , so $50000 \leq 198 k<60000 \Rightarrow k=7 \times 37$.
Hence $n=198 k=51282$ and $m=28215 \Rightarrow m+n=79497$.