Quick Solution

This is a permutation problem. We can solve it by considering the number of available choices for each of the three digit positions (hundreds, tens, and units), making sure to apply the given constraints.

1. Analyze the Constraints

• Digits available: {2, 3, 5, 6, 7, 9} (a total of 6 digits).
• The number must be divisible by 5. This means the last digit (the units place) must be 5, as 0 is not available.
• There is no repetition of digits, so each digit can be used only once.

2. Fill the Digit Slots

Let's think of the 3-digit number as three empty slots to fill: [Hundreds] [Tens] [Units]. It's best to fill the most restricted slot first.

1. Units Place: Because the number must be divisible by 5, this digit must be the '5' from our list. There is only 1 choice for this position.
   [__] [__] [5]
2. Hundreds Place: We started with 6 available digits. Since we have already used the digit '5' for the units place, there are 5 remaining choices for the hundreds place.
   [5 choices] [__] [1 choice]
3. Tens Place: We have now used two distinct digits (one for the hundreds place and '5' for the units place). This leaves 4 remaining choices for the middle (tens) slot.
   [5 choices] [4 choices] [1 choice]

3. Calculate the Total Number of Possibilities

The total number of 3-digit numbers that can be formed is the product of the number of choices for each position.

Total Numbers = 5 × 4 × 1 = 20

Final Answer: 20 such numbers can be formed.