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 96 p is equal to ________ .

Answer 1

The correct answer is 33
Total number of numbers from given is taken from the following condition
Condition = n(s) = 26.
Every required number is of the form
$A = 7.(10^{a1} + 10^{a2} + 10^{a3} + ...... ) + 111111$
Here 111111 is always divisible by 21.
Therefore, If A is divisible by 21 then
$10^{a1} \+ 10^{a2} \+ 10^{a3} \+ ......$
must be divisible by 3.
For this we have
$^6C_0 + ^6C_3 + ^6C_6$
cases are there
$∴ n(E) = ^6C_0 + ^6C_3 + ^6C_6 = 22$
∴ Required probability
$= \frac{22}{2^6}= p$

$∴ \frac{11}{32} = p$
Therefore , 96 p = 33