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(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. Let the four-digit number be $d_1 d_2 d_3 d_4$. There are $9$ choices for

Vedclass · Accepted Answer

$2187$

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(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. Let the four-digit number be $d_1 d_2 d_3 d_4$. There are $9$ choices for each of the first three digits $(d_1, d_2, d_3)$,which are ${1, 2, 3, 4, 5, 6, 7, 8, 9}$. Total ways to choose the first three digits is $9 	imes 9 	imes 9 = 729$. Let the sum of the first three digits be $S = d_1 + d_2 + d_3$. For the whole number to be divisible by $3$,$S + d_4$ must be a multiple of $3$. For any value of $S$,we check the possible values of $d_4 \in {1, 2, 3, 4, 5, 6, 7, 8, 9}$ such that $S + d_4 \equiv 0 \pmod{3}$. If $S \equiv 0 \pmod{3}$,then $d_4 \in {3, 6, 9}$ ($3$ choices). If $S \equiv 1 \pmod{3}$,then $d_4 \in {2, 5, 8}$ ($3$ choices). If $S \equiv 2 \pmod{3}$,then $d_4 \in {1, 4, 7}$ ($3$ choices). In all cases,there are exactly $3$ choices for $d_4$. Therefore,the total number of such four-digit numbers is $729 	imes 3 = 2187$.

The number of four-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ which are divisible by $3$,when repetition of digits is allowed any number of times,is

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