3 digit numbers are constructed from the digits 0-9, using each digit at most once. How many numbers: end in 0

We are tasked with finding how many 3-digit numbers can be constructed from the digits 0-9, using each digit at most once, with the condition that the number ends in 0.

### Steps:

1. **Fix the last digit**: 
   Since the number must end in 0, the last digit is fixed as 0.

2. **Choose the first digit**: 
   The first digit must be a non-zero digit (since it’s a 3-digit number), and it must be chosen from the digits 1-9. So, there are 9 possible choices for the first digit.

3. **Choose the second digit**: 
   After the first digit is chosen, one digit has been used up. Therefore, there are 8 remaining digits to choose from for the second digit.

### Total number of possibilities:
- First digit: 9 choices (1-9),
- Second digit: 8 choices (remaining digits after choosing the first),
- Last digit: fixed as 0.

Thus, the total number of 3-digit numbers that end in 0 is:

\[
9 \times 8 = 72
\]

Therefore, **72 numbers** can be formed where the 3-digit number ends in 0.

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### Related questions:
1. How many 3-digit numbers can be formed without any restrictions?
2. How many 3-digit numbers can be formed that start with 0?
3. How many 3-digit numbers can be formed that do not contain the digit 0?
4. How many 4-digit numbers can be formed with distinct digits?
5. How many 3-digit numbers can be formed that end in a specific digit, say 5?

### Tip:  
When solving problems involving digits, always check if the placement of 0 is restricted since it cannot be the first digit of a multi-digit number.

3 digit numbers are constructed from the digits 0-9, using each digit at most once. How many numbers end in 0?

Find out how many 3-digit numbers ending in 0 can be constructed using the digits 0-9, with each digit used at most once. This problem introduces key combinatorics concepts, including choosing digits and using the fundamental counting principle. Suitable for students in Grades 6-8.

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