Key Concepts and Formulas

1. Probability Definition: The probability of an event $E$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes: $P(E) = \frac{\text{Number of favorable outcomes (n(E))}}{\text{Total number of possible outcomes (n(S))}}$
2. Combinatorics: Used to count the total number of possible arrangements (numbers) and the number of specific arrangements (favorable outcomes). For choosing $k$ items from a set of $n$ items without regard to the order, the combination formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
3. Divisibility Rules:
   • Divisibility by 21: A number is divisible by 21 if and only if it is divisible by both 3 and 7, since 3 and 7 are coprime prime numbers.
   • Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
   • Divisibility by 7: We will use modular arithmetic to efficiently check this condition.
4. Modular Arithmetic: The notation $a \equiv b \pmod m$ means that $a$ and $b$ have the same remainder when divided by $m$. This is particularly useful for analyzing divisibility.

Step-by-Step Solution

Step 1: Determine the Total Number of Possible Outcomes (n(S))

• Understanding the structure: We need to form a 6-digit number using only digits 1 and 8. This means there are six positions, and for each position, we have 2 choices (1 or 8).
• Calculation: Since each digit choice is independent, we use the Multiplication Principle: $n(S) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64$ There are 64 distinct 6-digit numbers that can be formed using only digits 1 and 8.

Step 2: Determine the Number of Favorable Outcomes (n(E))

A number is a "favorable outcome" if it is a multiple of 21, meaning it must be divisible by both 3 and 7.

Step 2a: Condition for Divisibility by 7

Let the 6-digit number be $N = d_1 d_2 d_3 d_4 d_5 d_6$, where each digit $d_i \in \{1, 8\}$.

• Modular property of digits: Observe the digits 1 and 8 modulo 7:
  • $1 \equiv 1 \pmod 7$
  • $8 \equiv 1 \pmod 7$ Thus, every digit $d_i$ in our number is congruent to 1 modulo 7.
• Applying modular arithmetic to N: The number $N$ can be expressed as $N = d_1 \cdot 10^5 + d_2 \cdot 10^4 + d_3 \cdot 10^3 + d_4 \cdot 10^2 + d_5 \cdot 10^1 + d_6 \cdot 10^0$. Substituting $d_i \equiv 1 \pmod 7$ for each digit: $N \equiv (1 \cdot 10^5 + 1 \cdot 10^4 + 1 \cdot 10^3 + 1 \cdot 10^2 + 1 \cdot 10^1 + 1 \cdot 10^0) \pmod 7$ $N \equiv (10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0) \pmod 7$ The sum $10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0$ is the number $111111$. Now, we check $111111 \pmod 7$: $111111 = 7 \times 15873$. So, $111111 \equiv 0 \pmod 7$.
• Conclusion for divisibility by 7: Since $N \equiv 111111 \pmod 7$ and $111111 \equiv 0 \pmod 7$, it follows that $N \equiv 0 \pmod 7$. This means every 6-digit number formed using only digits 1 and 8 is automatically divisible by 7. Therefore, for a number to be a multiple of 21, we only need to ensure it is divisible by 3.

Step 2b: Condition for Divisibility by 3

Let $k$ be the number of times the digit '8' appears in the 6-digit number. The number of times '1' appears will be $(6-k)$. The value of $k$ can range from 0 to 6.

• Sum of digits (S): $S = (k \times 8) + ((6-k) \times 1)$ $S = 8k + 6 - k$ $S = 7k + 6$
• Applying divisibility by 3 condition: For $N$ to be divisible by 3, $S$ must be divisible by 3. $7k + 6 \equiv 0 \pmod 3$
• Simplifying modulo 3: Since $7 \equiv 1 \pmod 3$ and $6 \equiv 0 \pmod 3$: $1k + 0 \equiv 0 \pmod 3$ $k \equiv 0 \pmod 3$
• Possible values for k: The number of eights ($k$) must be a multiple of 3. Given $0 \le k \le 6$, the possible values for $k$ are:
  • $k=0$ (zero 8s, six 1s)
  • $k=3$ (three 8s, three 1s)
  • $k=6$ (six 8s, zero 1s)

Step 2c: Counting Favorable Outcomes (n(E))

We count the number of ways to form a 6-digit number for each valid $k$ using combinations:

• Case 1: $k=0$ (Number is 111111) Number of ways = $\binom{6}{0} = 1$.
• Case 2: $k=3$ (Three 8s and three 1s) Number of ways = $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
• Case 3: $k=6$ (Number is 888888) Number of ways = $\binom{6}{6} = 1$.
• Total Favorable Outcomes: $n(E) = 1 + 20 + 1 = 22$

Step 3: Calculate the Probability ($p$)

Using the formula $p = \frac{n(E)}{n(S)}$: $p = \frac{22}{64} = \frac{11}{32}$

Step 4: Calculate $96p$

The problem asks for the value of $96p$: $96p = 96 \times \frac{11}{32}$ $96p = (3 \times 32) \times \frac{11}{32}$ $96p = 3 \times 11$ $96p = 33$

Common Mistakes & Tips

• Misinterpreting "Divisible by 21": Always break down divisibility by composite numbers into their prime factors (e.g., $21 = 3 \times 7$).
• Forgetting to check all divisibility conditions: Ensure both (or all) conditions are met. In this problem, the divisibility by 7 condition was met by all numbers, which is a key simplification.
• Incorrectly applying combinatorics: Be careful when calculating combinations, especially for edge cases like $\binom{n}{0}$ and $\binom{n}{n}$.
• Calculation errors: Double-check arithmetic, especially with modular arithmetic and fractions.

Summary

We first determined the total number of 6-digit numbers formed by digits 1 and 8 as $2^6 = 64$. Next, we established that all such numbers are divisible by 7 using modular arithmetic (as $d_i \equiv 1 \pmod 7$ and $111111 \equiv 0 \pmod 7$). Then, we applied the divisibility rule for 3 to the sum of digits, finding that the number of '8's ($k$) must be a multiple of 3. Counting the arrangements for $k=0, 3, 6$ using combinations, we found 22 favorable outcomes. The probability $p$ was then $\frac{22}{64} = \frac{11}{32}$. Finally, $96p = 96 \times \frac{11}{32} = 33$.

The final answer is $\boxed{33}$.