Question

A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:

• A. 122
• B. 210
• C. 216
• D. 217

Answer 1

The Correct Option is C

To find the total number of ways to form a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition, we follow these steps:

1. A number is divisible by 3 if the sum of its digits is divisible by 3. For the digits 0, 1, 2, 3, 4, and 5, their total sum is 15, which is divisible by 3. Any five digits chosen from this set must also have a sum divisible by 3 to form a valid number.

2. The possible five-digit combinations are based on choosing any 5 out of the 6 digits (0, 1, 2, 3, 4, 5). Calculate combinations: C(6,5) = 6.

3. After choosing the five digits, ensure that their sum is divisible by 3. Since the sum of all six digits (15) is divisible by 3, leaving out any one digit will always result in a sum divisible by 3.

4. Next, distinguish two cases:

  a. The number begins with 0—this is invalid.

  b. The number does not begin with 0—this is valid.

5. Calculate permutations for each valid selection. When 0 is not among the digits:

  a. First digit is one of the 5 non-zero digits: 4 choices.

  b. Arrange the remaining 4 digits: 4! permutations.

6. Total valid permutations for numbers not starting with 0: 5 × 4!

7. Calculate: 5 × 24 = 120.

8. Repeat the same process if 0 is included:

  a. Choose 4 from 1, 2, 3, 4, 5: C(5,4) = 5 choices.

  b. Apply the similar counting for permutations: 4! × 4 = 96 (since 0 cannot be first).

9. Add both scenarios: 120 + 96 = 216.

Thus, the total number of ways is 216.