JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Permutations and Combinations

A number is divisible by $6$ when it is divisible by both $2$ and $3$. The only even digit available is $4$, so the number must end in $4$. The remaining five places are filled freely from $\{4,5,9\}$, whose residues modulo $3$ are $4\equiv 1$, $5\equiv 2$, $9\equiv 0$. With the fixed last digit contributing residue $1$, divisibility by $3$ requires the five free digits to have residue sum $\equiv 2(\mathrm{mod}3)$. Each place supplies exactly one digit of each residue $0,1,2$, so among the $3^5=243$ choices the three possible residue sums occur equally often, $\frac{243}{3}=81$ times each. Therefore the number of such six-digit numbers is $81$.