JEE 2023 — Mathematics PYQ

Solution:

Divisibility by 6: The number must be divisible by $2$ and $3$.

Divisibility by 2: The last digit must be $4$ (since $5,9$ are odd).

Divisibility by 3: The sum of all six digits must be a multiple of $3$.

Let the digits be $d_1,d_2,d_3,d_4,d_5$ and $d_6=4$.

Since $9$ is a multiple of $3$, we only care about the count of $4$s and $5$s.

Total sum $\equiv (n_4\times 4+n_5\times 5)(\mathrm{mod}3)\equiv (n_4+2n_5)(\mathrm{mod}3)$.

Analysis of combinations leads to the total count: 81.