JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Permutations and Combinations

Divisibility by $15$ requires divisibility by both $5$ and $3$. The only available digit for the units place is $5$, so the number is $d_1{d}_2{d}_{3}5$ with each $d_i\in \{1,2,3,5\}$, and the digit sum must be a multiple of $3$, i.e. $d_1+d_2+d_3\equiv 1(\mathrm{mod}3)$. Modulo $3$ the digit $3$ gives residue $0$, the digit $1$ gives $1$, and the digits $2,5$ give $2$, so each position supplies one digit of residue $0$, one of residue $1$ and two of residue $2$. Counting ordered residue-triples summing to $1(\mathrm{mod}3)$, weighted by these multiplicities, gives $\frac{4^3-1}{3}=21$ admissible numbers.