1. Total number of 6-digit numbers using only digits 1 and 8

Each of the 6 positions can be filled with either $1$ or $8$. So total numbers: $2^6 = 64$

2. Condition for divisibility by 21

Since $21 = 3 \times 7$ and $3,7$ are coprime, the number must be divisible by both $3$ and $7$.

3. Divisibility by 3

Let the number contain $k$ digits equal to $8$. Then it contains $(6-k)$ digits equal to $1$.

Sum of digits: $8k + 1(6-k) = 7k+6$ For divisibility by $3$: $7k+6 \equiv 0 \pmod{3}$ Since $7 \equiv 1 \pmod{3}$ and $6 \equiv 0 \pmod{3}$, $k \equiv 0 \pmod{3}$ So $k=0,3,6$

4. Check divisibility by 7

We now test numbers with $k=0,3,6$ eights.

Case 1: $k=0$

Number is $111111$ Now, $111111 = 7 \times 15873$ So it is divisible by $7$.

Case 2: $k=6$

Number is $888888 = 8 \times 111111$ Since $111111$ is divisible by $7$, $888888$ is also divisible by $7$.

Case 3: $k=3$

We need 6-digit numbers with exactly three $8$'s and three $1$'s. Total such numbers: $\binom{6}{3}=20$

Let us write any such number as: $N = 111111 + 7\times S$ where each position having $8$ instead of $1$ contributes $7\cdot 10^r$ for some place value $10^r$.

Since $111111$ is divisible by $7$, divisibility of $N$ by $7$ depends on $S = 10^{a}+10^{b}+10^{c}$ for the three positions where digit $8$ occurs.

Now modulo $7$: $10 \equiv 3 \pmod{7}$ so powers cycle as