Using the digits $0,1,2, \ldots 8,9$, determine how many 3-digit numbers can be constructed according to the following criteria.
The number must be odd and greater than 600 ; digits may be repeated.
Answer
Final Answer: The total number of 3-digit numbers that can be formed according to the given criteria is \(\boxed{150}\).
Step 1 ：Given the problem, we need to find out how many 3-digit numbers can be formed using the digits 0 to 9, with the conditions that the number must be odd and greater than 600, and digits may be repeated.
Step 2 ：Firstly, since the number must be odd, the units place can only be filled by the digits 1, 3, 5, 7, 9. This gives us 5 options for the units place.
Step 3 ：Secondly, the hundreds place must be filled by a digit greater than 6, so the options are 7, 8, 9. This gives us 3 options for the hundreds place.
Step 4 ：Thirdly, the tens place can be filled by any of the 10 digits (0-9), so we have 10 options for the tens place.
Step 5 ：Since the digits can be repeated, the total number of 3-digit numbers that can be formed is the product of the number of options for each place, which is \(3 \times 10 \times 5\).
Step 6 ：Calculating the above expression, we get \(3 \times 10 \times 5 = 150\).
Step 7 ：Final Answer: The total number of 3-digit numbers that can be formed according to the given criteria is \(\boxed{150}\).
