Question
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

Answer

a
(a)

As $N$ be the least positive integer and when a non-zero $\operatorname{digit} C$ is written af ter the last digit of $N$, the resulting number is divisible by $C$.

So, $10 N+C$ is divisible by $C$

$\therefore 10 N$ 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

$2+5+2=9$

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