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?

I mostly agree with the previous answer. You can add up all seven numbers in that set which is equal $23$. For a five digit number to be divisible by $3$ the sum of its digits has to be a multiple of three.
So we have to find all possible combinations of $5$ digit numbers that when all digits are added equals a multiple of $3$.
The sum of the $7$ numbers equals $23$ and we have to subtract $23$ by two numbers from the set to get the sum of five numbers. $23 - a - b$
Where $a+b$ has to equal a multiple of $3 +2$. e.g. $3n+2$. So like $5$ or $8$. Because $23 - (3n+2) = 21 - 3n$ which is divisible by $3$ meaning the five digit number is divisible by $3$.
So we find all possible pairs that equals a multiple of $3 + 2$
$(0,2)  (1,4)  (1,7)  (2,3)  (2,6)  (4,7)$
Which is similar to the previous answer except there are more pairs. There are $5!$ or 120 ways to order $5$ digits so the answer is $6×120=720$
This is true assuming that $02367$ is considered to be a $5$ digit number. It might not be considered a $5$ digit number since it starts with zero. Otherwise the answer would be $720 - 5×4! = 600$
