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(A) Let the number be $N$. When a non-zero digit $c$ is appended to $N$,the new number is $10N + c$.Given that $10N + c$ is divisible by $c$ for all $

Vedclass · Accepted Answer

$9$

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(A) Let the number be $N$. When a non-zero digit $c$ is appended to $N$,the new number is $10N + c$.Given that $10N + c$ is divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.This implies that $10N$ must be divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.Therefore,$10N$ must be a multiple of the least common multiple $(LCM)$ of the digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.$LCM(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 	imes 3^2 	imes 5 	imes 7 = 8 	imes 9 	imes 5 	imes 7 = 2520$.So,$10N$ must be a multiple of $2520$,which means $N$ must be a multiple of $252$.The least positive integer $N$ is $252$.The sum of the digits of $N$ is $2 + 5 + 2 = 9$.

Let $N$ be the least positive integer such that whenever a non-zero digit $c \in \{1, 2, \dots, 9\}$ is written after the last digit of $N$,the resulting number is divisible by $c$. The sum of the digits of $N$ is:

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