The total number of three-digit numbers,divis | vedclass

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(A) three-digit number is divisible by $3$ if the sum of its digits is divisible by $3$.Let the digits be $d_1, d_2, d_3 \in \{1, 3, 5, 8\}$.The sum $

Vedclass · Accepted Answer

$22$

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(A) three-digit number is divisible by $3$ if the sum of its digits is divisible by $3$.Let the digits be $d_1, d_2, d_3 \in \{1, 3, 5, 8\}$.The sum $S = d_1 + d_2 + d_3$ must be a multiple of $3$.We analyze the digits modulo $3$: $1 \equiv 1, 3 \equiv 0, 5 \equiv 2, 8 \equiv 2$.Let $n_0, n_1, n_2$ be the number of digits in the set congruent to $0, 1, 2 \pmod 3$ respectively.Here,$n_0 = 1$ (digit $3$),$n_1 = 1$ (digit $1$),$n_2 = 2$ (digits $5, 8$).Using the generating function method or counting cases based on the sum modulo $3$,the total number of such sequences is given by $\frac{1}{3}(4^3 + 2 	imes (n_0 + \omega n_1 + \omega^2 n_2)^3)$ where $\omega$ is the cube root of unity,or simply by checking combinations.The number of combinations $(d_1, d_2, d_3)$ such that $d_1+d_2+d_3 \equiv 0 \pmod 3$ is $22$.

The total number of three-digit numbers,divisible by $3$,which can be formed using the digits $1, 3, 5, 8$,if repetition of digits is allowed,is:

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