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(A) Let $n$ be an integer. When a nonzero digit $d$ is appended to the right of $n$,the new number is $10n + d$.We are given that $10n + d$ is divisib

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$20$

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(A) Let $n$ be an integer. When a nonzero digit $d$ is appended to the right of $n$,the new number is $10n + d$.We are given that $10n + d$ is divisible by $d$ for all $d \in \{1, 2, \dots, 9\}$.This implies $\frac{10n + d}{d} = \frac{10n}{d} + 1$ must be an integer for all $d \in \{1, 2, \dots, 9\}$.Thus,$10n$ must be divisible by every $d \in \{1, 2, \dots, 9\}$.This means $10n$ must be a multiple of the least common multiple of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.$	ext{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 	imes 3^2 	imes 5 	imes 7 = 2520$.So,$10n$ must be a multiple of $2520$,which means $n$ must be a multiple of $252$.The smallest positive integer $n$ is $252$.The product of the digits of $N = 252$ is $2 	imes 5 	imes 2 = 20$.

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