4-digit numbers are formed using the digits 4 | vedclass

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(D) The total number of $4$-digit numbers that can be formed using the digits ${4, 5, 6, 7, 8, 9}$ with repetition is $6^4 = 1296$. Let the number be

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) The total number of $4$-digit numbers that can be formed using the digits ${4, 5, 6, 7, 8, 9}$ with repetition is $6^4 = 1296$. Let the number be $N = d_1 d_2 d_3 d_4$. The sum of digits is $S = d_1 + d_2 + d_3 + d_4$. $A$ number is divisible by $3$ if and only if the sum of its digits is divisible by $3$. Consider the digits modulo $3$: $4 \equiv 1 \pmod{3}$, $5 \equiv 2 \pmod{3}$, $6 \equiv 0 \pmod{3}$, $7 \equiv 1 \pmod{3}$, $8 \equiv 2 \pmod{3}$, $9 \equiv 0 \pmod{3}$. There are two digits for each remainder $0, 1, 2$ modulo $3$. Let $S_n$ be the sum of the first $n$ digits. For any choice of $d_1, d_2, d_3$, the sum $S_3 = d_1 + d_2 + d_3$ will have some remainder $r \in \{0, 1, 2\}$ modulo $3$. To make $S_4 = S_3 + d_4$ divisible by $3$, we need $d_4 \equiv -S_3 \pmod{3}$. Since there are exactly two choices for $d_4$ for each remainder, and there are $6$ total choices for $d_4$, the probability that $d_4$ satisfies the condition is $\frac{2}{6} = \frac{1}{3}$. Thus, the probability is $\frac{1}{3}$.

$4$-digit numbers are formed using the digits $4, 5, 6, 7, 8, 9$ allowing repetition of the given digits. If a number is chosen at random from those numbers thus formed, then the probability that it is exactly divisible by $3$ is

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