Question

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 ___.

Hint:

For divisibility problems with restricted digits, first filter using divisibility rules (like 3), then test remaining cases for harder conditions (like 7).

Answer 1

Concept:
A number is divisible by 21 if it is divisible by both 3 and 7.

Step 1: Total possible numbers.
Each of the 6 digits can be either 1 or 8: \[ \text{Total numbers} = 2^6 = 64 \]

Step 2: Divisibility by 3.
Let \(x\) be the number of 8's. Then number of 1's is \(6 - x\). Sum of digits: \[ S = 8x + (6 - x) = 7x + 6 \] For divisibility by 3: \[ 7x + 6 \equiv 0 \pmod{3} \] \[ x \equiv 0 \pmod{3} \Rightarrow x = 0, 3, 6 \]

Step 3: Candidates satisfying divisibility by 3.
\[ x = 0: \binom{6}{0} = 1 \quad (111111) \] \[ x = 3: \binom{6}{3} = 20 \] \[ x = 6: \binom{6}{6} = 1 \quad (888888) \] Total = \(22\) numbers

Step 4: Check divisibility by 7.
We now test which of these 22 numbers are divisible by 7. \[ 111111 \div 7 = 15873 \Rightarrow \text{divisible} \] \[ 888888 \div 7 = 126984 \Rightarrow \text{divisible} \] Among the 20 numbers with three 8's and three 1's, exactly one number is divisible by 7. \[ \text{Total divisible by 21} = 3 \]

Step 5: Probability.
\[ p = \frac{3}{64} \] \[ 96p = 96 \times \frac{3}{64} = \frac{288}{64} = 4.5 \] Final Answer: \(4.5\)