To find the number of three-digit numbers divisible by 5 with no repeated digits, let the three-digit number be represented as ABC, where A, B, and C are digits.

We have the following conditions:

Three-digit number: A must be a digit from 1 to 9 (A ∈ {1, 2, …, 9}). B and C can be any digit from 0 to 9 (B, C ∈ {0, 1, …, 9}).
Divisible by 5: The last digit C must be either 0 or 5.
No repeated digits: A ≠ B, B ≠ C, and A ≠ C.

We can analyze this problem by considering the two possible values for C.

Case 1: The last digit C is 0.

If C = 0, the number is of the form AB0.

For C: There is only 1 choice (0).
For A: A cannot be 0 (Property 1, three-digit number). Also, A cannot be C (Property 3, no repeated digits), so A cannot be 0. Thus, A can be any digit from {1, 2, 3, 4, 5, 6, 7, 8, 9}. This gives 9 choices for A.
For B: B cannot be C (Property 3), so B cannot be 0. B also cannot be A (Property 3). Since A is one of the 9 non-zero digits, B must be chosen from the remaining 8 non-zero digits. (e.g., if A=1, B can be any of 2,3,4,5,6,7,8,9). This gives 8 choices for B.

Total numbers for Case 1 = (Choices for A) × (Choices for B) × (Choices for C)

Total = 9 × 8 × 1 = 72.

Case 2: The last digit C is 5.

If C = 5, the number is of the form AB5.

For C: There is only 1 choice (5).
For A: A cannot be 0 (Property 1). A also cannot be C (Property 3), so A cannot be 5.

Therefore, A can be any digit from {1, 2, 3, 4, 6, 7, 8, 9}. This gives 8 choices for A.
For B: B cannot be C (Property 3), so B cannot be 5. B also cannot be A (Property 3).

The available digits for B are all digits from 0 to 9, excluding C (5) and excluding A.

Since there are 10 total digits, excluding 5 leaves 9 digits. Excluding A (which is one of the 8 choices for A) leaves 8 digits. So, B has 8 choices.

(e.g., if A=1, B can be any of 0, 2, 3, 4, 6, 7, 8, 9).

Total numbers for Case 2 = (Choices for A) × (Choices for B) × (Choices for C)

Total = 8 × 8 × 1 = 64.

Total Number of such three-digit numbers

Adding the numbers from both cases:

Total = 72 (from Case 1) + 64 (from Case 2) = 136.

Thus, there are 136 three-digit numbers divisible by 5 in which no digits are repeated.

The final answer is $\boxed{\text{136}}$.

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Permutations and Combinations · Mathematics · IAT (IISER)

MCQ (Single Correct Answer)