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 numbersAdding 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 .