Number of 4 digited numbers using 1, 2, 3, 4, 5, 6 divisible by 3 is (with repetition)

Given digits 1, 2, 3, 4, 5, 6 (consecutive digits)

Total number of numbers $=6^4$

Among these total numbers, some numbers are divisible by 3

By fixing first 3 places with some fixed digits from 1, 2, 3, 4, 5, 6 we get the following 6 consecutive numbers

$\begin{array}{c}\begin{array}{cccc}{a}_1& a_2& a_3& 1\end{array}\\ \begin{array}{cccc}{a}_1& a_2& a_3& 2\end{array}\\ \begin{array}{cccc}{a}_1& a_2& a_3& 3\end{array}\\ \begin{array}{cccc}{a}_1& a_2& a_3& 4\end{array}\\ \begin{array}{cccc}{a}_1& a_2& a_3& 5\end{array}\\ \begin{array}{cccc}{a}_1& a_2& a_3& 6\end{array}\end{array}$

We know that out of any 6 consecutive integers exactly 2 are divisible by 3

$\therefore$ $\frac{1}{3}rd$  of these $6^4$  numbers are divisible by 3

$\therefore$  Required number $=\frac{1}{3}\times 6^4=2\times 6^3=432$