If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is $$p$$, then $$96p$$ is equal to ______

We need to find the probability that a randomly chosen 6-digit number formed using only digits 1 and 8 is a multiple of 21, and then compute $$96p$$.

First, each of the 6 positions can be either 1 or 8, so the total number of such 6-digit numbers is $$2^6 = 64$$.

Next, a 6-digit number with digits $$d_1d_2d_3d_4d_5d_6$$ where each $$d_i \in \{1,8\}$$ can be written as

$$N = \sum_{i=1}^{6} d_i \cdot 10^{6-i}\;.$$

Now we set $$d_i = 1 + 7b_i$$ where $$b_i \in \{0,1\}$$ to obtain

$$N = \sum_{i=1}^{6}(1 + 7b_i)\cdot 10^{6-i} = 111111 + 7\sum_{i=1}^{6} b_i \cdot 10^{6-i}\;.$$

Now we check divisibility by 21, noting that $$21 = 3 \times 7$$. For divisibility by 3, the digit sum must be divisible by 3. If $$k$$ of the digits are 8 and the remaining $$(6-k)$$ are 1, then the digit sum is

$$8k + (6-k) = 6 + 7k\;.$$

Since

$$6 + 7k \equiv 0 \pmod{3} \iff 7k \equiv 0 \pmod{3} \iff k \equiv 0 \pmod{3}\;,$$

we conclude that $$k \in \{0,3,6\}$$.

For divisibility by 7, observe that

$$N = 111111 + 7M\quad\text{where}\quad M = \sum_{i=1}^6 b_i\cdot 10^{6-i}\;,$$

and since $$111111 = 7 \times 15873\,$$ it follows that $$N$$ is always divisible by 7.

Therefore, the favorable outcomes occur when $$k \in \{0,3,6\}$$, giving

$$\binom{6}{0} + \binom{6}{3} + \binom{6}{6} = 1 + 20 + 1 = 22\;.$$

Finally, the required probability is

$$p = \frac{22}{64} = \frac{11}{32}\;,$$

and hence

$$96p = 96 \times \frac{11}{32} = 3 \times 11 = 33\;.$$

Hence the answer is $$33$$.