Question

The digits $1,2,3,4,5,6,7,8$ and $9$ are written in random order to form a nine digit number. Let $p$ be the probability that this number is divisible by $36$, find $9p$

Answer 1

Since $1+2+...+9=36$, a number consisting all these digits will be divisible by $9$. Thus, the number will be divisible by $36$ if and only if it is divisible by $4$.
There are ${}^9{P}_9=9!$ ways of arranging the nine digits to form a nine-digit number. If this number is to be divisible by $4$, then its last digit must be even and the number formed by its last two digits must be divisible by $4$. This limits the possible values of the last pair to the following:
$12,32,52,72,92,$
$24,64,84,$
$16,36,56,76,96,$
$28,48,68.$
That is, there are $16$ ways of choosing the last two digits. The remaining digits can be arranged in ${}^7{P}_7=7!$ ways. Therefore the number of favorable arrangements is $(7!)\left(16\right)$ out of a total of $9!$, so that the required probability is
$\frac{\left(16\right)(7!)}{9!}=\frac{16}{9\times 8}=\frac{2}{9}$.