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
(a) 9
(b) 18
(c) 27
(d) 36
The correct option is (a) 9.
N is least positive integer and when a digit c is written after last digit of N, the resulting number is divisible by c.
⇒ So, (N)c / c
Least positive integer (N) which is divisible by digit c i.e. (1−9)
So possible number is
⇒ (9 × 8 × 7 × 5)c / c
Number, 9 × 8 × 7 × 5 is divisible by all the digits from 1 to 9
⇒ So, Number is 72 × 35 = 2520 = N
⇒ Sum of the digits of number N = 2 + 5 + 2 + 0 = 9
Answer is : (a) 9
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, 10N + 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