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 96 p is equal to _______________.

<p>Total number of numbers from given</p> <p>Condition = n(s) = 2⁶.</p> <p>Every required number is of the form</p> <p>A = 7 . (10^a₁ + 10^a₂ + 10^a₃ + .......) + 111111</p> <p>Here 111111 is always divisible by 21.</p> <p>$\therefore$ If A is divisible by 21 then</p> <p>10^a₁ + 10^a₂ + 10^a₃ + ....... must be divisible by 3.</p> <p>For this we have ⁶C₀ + ⁶C₃ + ⁶C₆ cases are there</p> <p>$\therefore$ n(E) = ⁶C₀ + ⁶C₃ + ⁶C₆ = 22</p> <p>$\therefore$ Required probability = ${{22} \over {{2^6}}} = p$</p> <p>$\therefore$ ${{11} \over {32}} = p$</p> <p>$\therefore$ $96p = 33$</p>