Number of 4 digited numbers using 0, 1, 2, 3, 4, 5 divisible by 6 with repetition is

Given digits 0,1, 2, 3, 4, 5 (consecutive digits)

Total number of numbers $=5\times 6^3$

Among these total numbers, some numbers are divisible by 6

By fixing first  places with some fixed digits from 1, 2, 3, 4, 5 and 2^nd ,3^rd place with some fixed digits 0,1,2,3,4,5 we get the following consecutive numbers

$\begin{array}{c}{a}_1{a}_2{a}_{3}0\\ a_1{a}_2{a}_{3}1\\ a_1{a}_2{a}_{3}2\\ a_1{a}_2{a}_{3}3\\ a_1{a}_2{a}_{3}4\\ a_1{a}_2{a}_{3}5\end{array}$

We know that out of any 6 consecutive integers exactly 1 are divisible by 6

$\therefore$ $\frac{1}{6}th$  of these $5\times 6^3$  numbers are divisible by 6

$\therefore$  Required number $=\frac{1}{6}\times 5\times 6^3=180$