2011/12 AHSMC Part 1 Problem 16

Note that if we write a positive integer $m$ in base $3$, then the base $3$ representation of $\bigl\lfloor \frac{m}{3} \bigr\rfloor$ is simply the base $3$ representation of $m$ with the rightmost digit removed. Also, a positive integer $m$ is divisible by $3$ if and only if the rightmost digit of $m$ is $0$. Hence, in order that none of $a_1$, $a_2$, $a_3$ and $a_4$ is divisible by $3$, the rightmost four digits of the base $3$ representation of $n$ are all non-zero. Note that $1000 > 2(3^5 + 3^4 + 3^3 + 3^2 + 3 + 1)$. If $n$ has at most $6$ digits in its base $3$ representation, the first two can be any of $0$, $1$ and $2$, while the last four cannot be $0$. There are $3^2 \times 2^4 = 144$ such numbers. Clearly, $n$ cannot have more than $7$ digits as otherwise $n \ge 3^7 > 1000$. Suppose $n$ has exactly $7$ digits. As before, the last four cannot be $0$. Since $1000 < 3^6 + 3^5 + 3^3 + 3^2 + 3 + 1$, the first one must be $1$, the second must be $0$ and the third can be any of $0$, $1$ and $2$. Hence, there are $3 \times 2^4 = 48$ such numbers. The total is $192$. The answer is (B).