To find how many 3-digit numbers can be made using only the digits $0, 1, 2, 3, 4, 5$, let's break this down step-by-step:

1. The hundreds place:

   • It cannot be 0 (since it's a 3-digit number).
   • So, the possible choices for the hundreds place are: $1, 2, 3, 4, 5$.
   • This gives us 5 possible choices for the hundreds digit.

2. The tens place:

   • Here, we can use any of the digits $0, 1, 2, 3, 4, 5$, meaning we have 6 possible choices.

3. The ones place:

   • Similarly, any of the digits $0, 1, 2, 3, 4, 5$ can be used here as well, giving us 6 possible choices.

Now, the total number of 3-digit numbers is the product of the number of choices for each place:

$5 \times 6 \times 6 = 180$

Therefore, 180 three-digit numbers can be made using the digits $0, 1, 2, 3, 4, 5$.

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Would you like more details or have any questions? Here are some related questions to explore:

1. How many 4-digit numbers can be formed using the same digits?
2. What if repetition of digits is not allowed?
3. How many 3-digit even numbers can be formed with these digits?
4. How many 3-digit numbers can be formed using only the digits $1, 2, 3$?
5. What if we are asked to form a number divisible by 5?

Tip: Always pay attention to restrictions on digit positions like the hundreds place, which cannot be zero in a multi-digit number.