Let N be the least positive integer such that whenever a non-zero digit c is written after the last digit of N, the resulting number is divisible by c. The sum of the digits of N is

As N be the least positive integer and when a non-zero digit C is written after the last digit of N, the resulting number is divisible by C. $\mathrm{So},10 \mathrm{N}+\mathrm{C}$ is divisible by C
$\therefore 10 \mathrm{N}$ must be divisible by C Now, the least integer (N) which is divisible by digit 'C i.e. (1 to 9) must be L.C.M of $\{1,3,4,6,7,9\}$
$\begin{array}{l}=\mathrm{L}.\mathrm{C}.\mathrm{M}\text{ of }\{4,7,9\}\\ =252=\mathrm{N}\end{array}$
and sum of digits of number ' N is
$2+5+2=9$