16. The digits from 1 to 9 are randomly arranged to make a 9 -digit number. What is the probability that the resulting number is divisible by 18 ?
A $\frac{1}{3}$
B $\frac{4}{9}$
C $\frac{1}{2}$
D $\frac{5}{9}$
$\mathrm{E} \frac{3}{4}$
SOLUTION
B

To be divisible by 18 , the number must be divisible by 2 and by 9 . The digit sum for any number formed is $1+2+3+4+5+6+7+8+9=45$, and hence every number is a multiple of 9 . Therefore any even number formed will be divisible by 18 . Of the 9 possible last digits, there are four even digits $(2,4,6,8)$, so the probability of the number being even is $\frac{4}{9}$.