How many five-digit numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without repeating any digit?120156196216

To solve this problem, we need to consider two cases:

1. The first digit is zero: In this case, we have 4 choices for the second digit (1, 2, 3, 4, or 5), 4 choices for the third digit, 3 choices for the fourth digit, and 2 choices for the fifth digit. However, we need to ensure that the sum of the digits is divisible by 3. Since 0 + 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 3, we need to exclude the cases where the sum of the digits is 1 or 2 mod 3. There are 4 such cases (120, 150, 210, 510), so the total number of numbers in this case is 443*2 - 4 = 92.

2. The first digit is not zero: In this case, we have 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the fifth digit. Again, we need to ensure that the sum of the digits is divisible by 3. There are 20 such cases (123, 132, 213, 231, 312, 321, 102, 120, 201, 210, 301, 310, 105, 150, 501, 510, 204, 240, 402, 420), so the total number of numbers in this case is 54321 - 20 = 100.

Therefore, the total number of five-digit numbers divisible by 3 that can be formed using the digits 0, 1, 2, 3, 4 and 5, without repeating any digit, is 92 + 100 = 192.