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

Step-by-step solution

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. So , 10   N + C is divisible by C ∴ 10   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 = L . C . M  of  4 , 7 , 9 = 252 = N and sum of digits of number ' N is   2 + 5 + 2 = 9