How many numbers can be formed using $1,3,4,5,6,9$ only once and divisible by $7$ or $11$ or $13$?
I know the divisibility rules but they are not helping here. how do i solve this? Also the question is unclear that do I need to use all of them or not.
My work -
I think if it was "$7$ and $11$ and $13$" then I would solve it by this -
any number in form $abcabc$ is divisible by $1001 = 7\times 13 \times 11$. Then the answer would be $\binom{6}{3}$? Correct me if i am wrong.
I will like to know both solutions. i.e. "$7$ or $11$ or $13$" and "$7$ and $11$ and $13$".
The six-digit numbers using the digits 1,3,4,5,6,9 that are divisible by $7$, $11$ or $13$ don't seem to satisfy any obvious pattern so I'm not sure there is a nice mathematical solution (maybe there is), but it's very easy for a computer to iterate over all permutations of the list 1,3,4,5,6,9 and test divisibility.
In this way you find the number of solutions divisible by $7$ OR $11$ OR $13$ is 180; and the number of solutions divisible by $7$ AND $11$ AND $13$ is 18.