The number of five-digit numbers that are div | vedclass

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(D) number is divisible by $6$ if it is even and the sum of its digits is divisible by $3$. Let the five-digit number be $d_1 d_2 d_3 d_4 d_5$. $d_1 \

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(D) number is divisible by $6$ if it is even and the sum of its digits is divisible by $3$. Let the five-digit number be $d_1 d_2 d_3 d_4 d_5$. $d_1 \in \{1, 2, 3, 4, 5\}$ ($5$ choices). $d_2, d_3 \in \{0, 1, 2, 3, 4, 5\}$ ($6$ choices each). $d_5 \in \{0, 2, 4\}$ ($3$ choices). Let $S = d_1 + d_2 + d_3 + d_4$. For the number to be divisible by $3$,$S + d_5 \equiv 0 \pmod{3}$. For any choice of $d_1, d_2, d_3$,the sum $d_1 + d_2 + d_3$ can be $0, 1, 	ext{ or } 2 \pmod{3}$. If $d_5 = 0$,$d_4$ must be chosen such that $d_1 + d_2 + d_3 + d_4 \equiv 0 \pmod{3}$. There are $2$ choices for $d_4$ in each case. If $d_5 = 2$,$d_4$ must be chosen such that $d_1 + d_2 + d_3 + d_4 \equiv 1 \pmod{3}$. There are $2$ choices for $d_4$ in each case. If $d_5 = 4$,$d_4$ must be chosen such that $d_1 + d_2 + d_3 + d_4 \equiv 2 \pmod{3}$. There are $2$ choices for $d_4$ in each case. Total numbers $= 5 	imes 6 	imes 6 	imes 2 	imes 3 = 1080$. Thus,option $(D)$ is correct.

The number of five-digit numbers that are divisible by $6$ which can be formed by choosing digits from $\{0, 1, 2, 3, 4, 5\}$,when repetition is allowed,is

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