One case is 49 dividing 59241 remainder is 000 and quotient is 1209

See the following program

```
#include <stdio.h>
main()
{
int i, j, k, l, m, number;
for (i=1; i<10; i++)
for (j=0; j<10; j++)
{
for (k=1; k<10; k++)
for (l=1; l<10; l++)
for (m=1; m<10; m++)
{
number = (10*i + j) * (1000*k+100*l+0+m);
if (number <100000)
{
if ((number%1000 >199) && (number%1000 <300))
{
if (((10*i + j)*(10*k + l)) != (number/100))
{
if (((number/1000) - ((10*i + j)*k)) > 9)
{
if (((10*i + j) * l) < 100)
printf ("%d %d * %d %d 0 %d = %d\n", i, j, k, l, m, number);
}
}
}
}
}
}
}
```

Catch is the middle digit should be 2, second digit in the quotient is zero and one more point...

But very good puzzle keep it up...