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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 $13$".

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    This isn't a "do your homework for free" service, what have you tried already? – 2017-01-12
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    "only once" usually means exactly once in problems like this. – 2017-01-12
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    @Travis added my work :) – 2017-01-12
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    I think the answer for "and" will be $^6p_3$ = 6*5*4. ?? – 2017-01-12
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    Number $abcabc$ uses each $a,$,$b$ and $c$ twice. You cant use numbers twice. – 2017-01-12
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    @JaroslawMatlak Opps. Then I am completely wrong :( – 2017-01-12
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    @JaroslawMatlak Then I think there is no solution for "and" right? – 2017-01-12
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    The divisibility test for $11$ is easy enough. It appears there are $2(3!)=72$ multiples of $11$. Unfortunately, the other $2$ divisibility tests aren't as easy and it could be a pain to find how many numbers are multiples of more than $1$ of the $3$ options. – 2017-01-12

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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.

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    Is that for using each number once? I believe the OP has determined there are no solutions for "OR". – 2017-01-12