We are asked to find the number of 5-digit odd numbers greater than 40,000 that can be formed using the digits \( 3, 4, 5, 6, 7, 0 \), such that at least one digit is repeated. To form a 5-digit number, we need to ensure that:
1. The number is odd, meaning the last digit must be one of \( 3, 5, 7 \).
2. The number must be greater than 40,000, so the first digit must be at least 4, i.e., \( 4, 5, 6, 7 \).
3. At least one digit must be repeated.
5
Step 1: Total number of 5-digit odd numbers greater than 40,000:
The first digit can be \( 4, 5, 6, \) or \( 7 \) (4 choices). The second, third, and fourth digits can be \( 0, 3, 4, 5, 6, 7 \) (6 choices each). The last digit can be \( 3, 5, 7 \) (3 choices). Thus, the total number of such 5-digit numbers is: \[ 4 \times 6 \times 6 \times 6 \times 3 = 2592. \]
Step 2: Now, we subtract the number of 5-digit odd numbers where no digits are repeated. The first digit can be chosen from \( 4, 5, 6, 7 \) (4 choices). The second digit can be chosen from \( 0, 3, 4, 5, 6, 7 \) excluding the first digit (5 choices).
The third digit can be chosen from the remaining 4 digits (4 choices). The fourth digit can be chosen from the remaining 3 digits (3 choices). The last digit can be chosen from \( 3, 5, 7 \), excluding the digits used already (2 choices). Thus, the number of such numbers is: \[ 4 \times 5 \times 4 \times 3 \times 2 = 480. \]
Step 3: Finally, subtract the number of numbers with no repeated digits from the total number of numbers: \[ 2592 - 480 = 2112. \] Therefore, the number of 5-digit odd numbers greater than 40,000, where at least one digit is repeated, is 2352.
For a non-zero complex number $ z $, let $\arg(z)$ denote the principal argument of $ z $, with $-\pi < \arg(z) \leq \pi$. Let $\omega$ be the cube root of unity for which $0 < \arg(\omega) < \pi$. Let $$ \alpha = \arg \left( \sum_{n=1}^{2025} (-\omega)^n \right). $$ Then the value of $\frac{3 \alpha}{\pi}$ is _____.