Step 1: Understanding the Concept:
A number is a multiple of 21 if it is a multiple of both 3 and 7. Since each digit is 1 or 8, we can analyze the divisibility rules for 3 and 7.
Step 2: Key Formula or Approach:
1. Total 6-digit numbers using 1 and 8: \( 2^6 = 64 \).
2. Divisibility by 3: Sum of digits must be a multiple of 3.
3. Divisibility by 7: Check using standard rules or remainders mod 7.
Step 3: Detailed Explanation:
Let \( n \) be the number of 8's in the 6-digit number. Then the number of 1's is \( 6-n \).
Sum of digits \( S = 8n + 1(6-n) = 7n + 6 \).
For \( S \) to be divisible by 3:
- If \( n=0 \), \( S=6 \) (Yes) \(\implies 111111\) (1 case)
- If \( n=3 \), \( S=27 \) (Yes) \(\implies \binom{6}{3} = 20\) cases
- If \( n=6 \), \( S=48 \) (Yes) \(\implies 888888\) (1 case)
Total numbers divisible by 3: \( 1 + 20 + 1 = 22 \).
Now, check divisibility by 7:
Note that \( 1 \equiv 1 \pmod 7 \) and \( 8 \equiv 1 \pmod 7 \).
Any 6-digit number \( d_5 d_4 d_3 d_2 d_1 d_0 \) mod 7 is:
\( \sum d_i 10^i \equiv \sum 1 \cdot 10^i \equiv 111,111 \pmod 7 \).
Since \( 111,111 = 7 \times 15873 \), it is always divisible by 7 regardless of the combination of 1s and 8s.
So all 22 numbers identified above are divisible by 21.
\( p = 22/64 \).
\( 96p = 96 \times (22/64) = 1.5 \times 22 = 33 \).
Step 4: Final Answer:
The value of 96p is 33.