<s>[INST] What is the number of 5 digit numbers divisible by 3? What is the number of $5$ digit numbers divisible by $3$ using the digits 
$0,1,2,3,4,6,7$ and repetition is not allowed?
 [/INST] the rule of three says that a number is divisible by three the sum of the digits is divisible by three.  SO if one five digit number is divisible by three then any arrangement of its digits is divisible by three.
So it's a matter of finding the ways of picking $5$ out of $7$ numbers that add to a multiple of $3$ and multiplying by the number of ways we can arrange $5$ digits.
If $0$ is among the $5$ the number can't start with $0$ so there are four choice for the fist digit and $4!$ for the rest so $4*4!$ total.  If $0$ is not among the $5$ there are $5!$ ways or arranging them.
Picking $5$ numbers that add to a multiple of three is a matter of picking the $2$ that we don't need.  $0+1+2 +3+4+6+7= 23\equiv 2 \mod 3$ so this is matter of picking two numbers that add to $2 \mod 3$ (so the remaining $5$ will add to $0 \mod 3$.  That is to say, $2$ that add to $2,5, 8,$ or $11$.
So those pairs are $(0,2),(1,4)(2,3),(1,7),(2,6),(4,7)$. 
Of the sets of five remaining, $(1,3,4,6,7), (0,2,3,6,7), etc.$ only one has no $0$; $(1,3,4,6,7)$ and $5!$ ways to arrange them.  The remaining $5$ have $0$s and so $4*4!$ ways to arange them.  So there are $5*(4*4!) + 5! = 4*5! +5! = 5*5! = 5*120 = 600$ possible numbers.
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