The condition for a number to be divisible by 3 is that the sum of individual digits must be divisible by 3. To form a five digit number (from the given numbers) which is divisible by 3 we can do it only with two set of numbers. They are {1,2,3,4,5} and {0,1,2,4,5}.

Case 1 : Using digits 1,2,3,4,5
In this case there is no restriction and the number of ways of arranging them is 5! = 120 ways

Case 2 : Using digits 0,1,2,4,5
In this case there is restriction that the first digit should not be zero.
The first place can be filled in 4 ways (any digit except 0). The second digit can be filled in 4 ways (any of the other 4 digits including 0). Similarly, the third, fourth, and fifth places can be filled in 3,2 and 1 ways respectively.
Number of ways = 4 x 4 x 3 x 2 x 1 = 96 ways

Total number of ways = 120 + 96 = 216 way

To find the number of five-digit numbers divisible by 3 using the given numerals 0, 1, 2, 3, 4, and 5 without repetition, we need to follow certain rules.

Rule 1: The sum of the digits of a number divisible by 3 must also be divisible by 3.

Rule 2: We cannot use the digit 0 as the leading digit of the number since it would make it a four-digit number.

Rule 3: The digit 0 can be used as any other digit in the number.

Now let's break down the solution into steps:

Step 1: Find the total number of ways to arrange the given numerals without repetition.

Since we need to form a five-digit number, the total number of ways to arrange the numerals is given by the permutation formula:

nPr = n! / (n-r)!

In this case, n = 6 (total number of numerals) and r = 5 (number of digits in the number).

Therefore, the total number of ways to arrange the numerals without repetition is:

6P5 = 6! / (6-5)! = 6! / 1! = 6! = 720

Step 2: Find the number of ways to arrange the numerals such that the sum of the digits is divisible by 3.

To find this, we need to consider the different cases where the sum of the digits can be divisible by 3.

Case 1: The sum of the digits is 3.

In this case, we have the following possibilities to form a five-digit number:

1, 2, 3, 4, 5
2, 1, 3, 4, 5
2, 3, 1, 4, 5
...
5, 4, 3, 2, 1

There are 5! = 120 ways to arrange these digits.

Case 2: The sum of the digits is 6.

In this case, we have the following possibilities to form a five-digit number:

1, 2, 3, 4, 6
1, 2, 3, 6, 4
1, 2, 6, 3, 4
...
6, 4, 3, 2, 1

Again, there are 5! = 120 ways to arrange these digits.

Case 3: The sum of the digits is 9.

In this case, we have the following possibilities to form a five-digit number:

1, 2, 3, 5, 4
1, 2, 3, 4, 5
1, 2, 5, 3, 4
...
4, 5, 3, 2, 1

Once again, there are 5! = 120 ways to arrange these digits.

Step 3: Calculate the total number of ways to arrange the numerals such that the sum of the digits is divisible by 3.

The total number of ways can be found by adding up the number of ways from each case:

Total number of ways =