Question

The number of $4$ digit numbers formed by $0,1,2,3,4,5$ (repetition of digits is allowed) such that it is divisible by $6$ is.

Answer 1

$X_1$ can be filled in $5$ ways whereas $X_2$ and $X_3$ can be filled in $6$ ways each.
For numbers to be divisible by $6$, $X_4$ must be even and $X_1+X_2+X_3+X_4=3k$ (where $k\in N$)
Now if $X_1+X_2+X_3=3k$, then $X_4=0$ only one way
If $X_1+X_2+X_3=3k+1$, then $X_4=2$ only one way
If $X_1+X_2+X_3=3k+2$, then $X_4=4$ only one way

$\therefore$ To satisfy above mentioned conditions $X_4$ can be filled in $1$ way
Hence the required number of numbers$=5\times 6\times 6\times 1=180$