How many three digit numbers can be formed that are divisible by 5 by using the

We need three-digit numbers divisible by 5 using digits 0,2,5,8,9 without repetition. Divisible by 5 means last digit is 0 or 5. Case 1: Last digit = 0. Remaining digits: {2,5,8,9}. First digit can't be 0 (already used), so 4 choices for first digit, then 3 for second. Total = 4 × 3 = 12. Case 2: Last digit = 5. Remaining digits: {0,2,8,9}. First digit cannot be 0. Total permutations of two digits from 4: 4P2 = 12. Subtract those with first digit 0: first digit fixed as 0, second digit any of remaining 3, so 3 invalid. Valid = 12 - 3 = 9. Alternatively, first digit from {2,8,9} (3 choices), second from remaining three (3 choices) = 9. Total = 12 + 9 = 21.