The number of numbers greater than 5000 and l | vedclass

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Question

(B) Let the $4$-digit number be $d_1 d_2 d_3 d_4$. Since the number is between $5000$ and $9000$, the first digit $d_1$ can be $5$ or $9$. Case $1$: $

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) Let the $4$-digit number be $d_1 d_2 d_3 d_4$. Since the number is between $5000$ and $9000$, the first digit $d_1$ can be $5$ or $9$. Case $1$: $d_1 = 5$. The sum of digits $S = 5 + d_2 + d_3 + d_4$ must be divisible by $3$. Possible values for $d_2, d_3, d_4 \in \{0, 1, 2, 5, 9\}$. There are $5 	imes 5 = 25$ possible combinations for $(d_2, d_3)$. For each pair, $d_4$ is determined such that $5 + d_2 + d_3 + d_4 \equiv 0 \pmod{3}$. If $5 + d_2 + d_3 \equiv 0 \pmod{3}$, then $d_4 \in \{0, 9\}$ ($2$ choices). If $5 + d_2 + d_3 \equiv 1 \pmod{3}$, then $d_4 \in \{2, 5\}$ ($2$ choices). If $5 + d_2 + d_3 \equiv 2 \pmod{3}$, then $d_4 \in \{1\}$ ($1$ choice). Counting the pairs $(d_2, d_3)$ by their sum modulo $3$: Sum $\equiv 0$: $(0,0), (0,9), (1,2), (1,5), (2,1), (2,4) 	ext{ (not in set)}, (5,1), (5,4) \dots$ Actually, checking all $25$ pairs: Sum $\equiv 0 \pmod{3}$: $9$ pairs. $9 	imes 2 = 18$ numbers. Sum $\equiv 1 \pmod{3}$: $8$ pairs. $8 	imes 2 = 16$ numbers. Sum $\equiv 2 \pmod{3}$: $8$ pairs. $8 	imes 1 = 8$ numbers. Total for $d_1=5$ is $18 + 16 + 8 = 42$. Wait, the condition is $d_1 Total numbers = $42$.

The number of numbers greater than $5000$ and less than $9000$ that are divisible by $3$, which can be formed using the digits $0, 1, 2, 5, 9$ with repetition allowed, is . . . . . . .

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