Dividing a $10$ digit number to get a answer between $1$ and $13$. I'd like to divide a $10$ digit number by any number between $2$ and $15$ and get any result between $1$ and $13$. That $10$ digit number has to contain the numbers from $0$ to $9$ and not repeat itself. (For example: $1,234,567,890$).
This $10$ digit number has to be divided by any number between $2$ and $15$ any number of times before finally getting an answer between $1$ and $14$. You cannot divide using a fraction nor get a fraction as an answer. An example of this is the $10$ digit number $2,143,968,750$. You can divide it by $15,\,15,\,15,\,15,\,14,\,11,\,11,\,5,\,5$ and get $1$ as an answer. I know there are about $35$ answers to this question but I'd like to find out how to get the answer.
Thank you.
 A: Since we're required to have a 10 digit number, there are $9*9!$ different 10 digit numbers using each of the numbers $0-9$ exactly once. This comes out to $3265920$ numbers. Of course, not all of these will be divisible by just the numbers below $15$.
Because 15 is not a prime, we don't actually need to test for divisibility by $15$, just at most $13$, and in fact setting the bounds to $16$ or $14$ would result in the exact same rules. For your example of $2143968750$, the simplest pattern of factorizations would be 2,3,3,3,3,5,5,5,5,5,5,7,11,11.
To create a number that has the highest chance of success, note a couple things: All of the 10-digit numbers created by these rules are divisible by $9$, and hence $3$ since the sum of the digits is 45, regardless of how you arrange them. If the last digit is 0, the number will be divisible by $10$, and hence $2$ and $5$. If the last three digits are divisible by $8$, then the entire number is divisible by 8 as well.
It would be relatively easy to create such a number programmatically by generating one of the 10-digit numbers and testing for divisibility, but there isn't a straight-forward, deterministic method for creating a number like that from scratch.
