JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Permutations and Combinations

Since the only usable digits are $0,1,2,5,9$ and the number must lie strictly between $5000$ and $9000$, the thousands digit must be $5$, so the number is $\overline{5abc}$ with $a,b,c\in \{0,1,2,5,9\}$. Divisibility by $3$ needs $5+a+b+c\equiv 0(\mathrm{mod}3)$, i.e. $a+b+c\equiv 1(\mathrm{mod}3)$. The residues modulo $3$ are $0\to \{0,9\}$, $1\to \{1\}$, $2\to \{2,5\}$, with counts $2,1,2$. Counting ordered triples $(a,b,c)$ whose residues sum to $1(\mathrm{mod}3)$ gives $42$. Hence the number of such numbers is $42$.