Question

If a five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, and 5 without repetition, then the total number of ways this can be done is:

• A. 120
• B. 144
• C. 192
• D. 216

Hint:

When forming numbers divisible by 3, check the sum of the digits and make sure the sum is divisible by 3. For five-digit numbers, be cautious when placing 0 as the first digit.

Answer 1

The Correct Option is D

We are tasked with forming a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition. A number is divisible by 3 if the sum of its digits is divisible by 3. Step 1: Total sum of the digits The sum of the digits 0, 1, 2, 3, 4, and 5 is: \[ 0 + 1 + 2 + 3 + 4 + 5 = 15. \] Since the total sum of the digits is 15, which is divisible by 3, the sum of the digits of any five-digit number formed from these digits will also be divisible by 3, provided that we leave out one of the digits. Step 2: Choosing one digit to leave out We can leave out any of the six digits (0, 1, 2, 3, 4, 5). If we leave out a digit, the sum of the remaining digits will still be divisible by 3. We need to calculate the number of five-digit numbers that can be formed with the remaining digits. Step 3: Counting the number of five-digit numbers We must choose 5 digits from the 6 available digits. We have 6 choices for the digit to leave out. For each selection of 5 digits, the number of ways to arrange them is given by the number of permutations of 5 digits. However, the first digit cannot be 0, so we need to adjust the counting. - If 0 is not selected, all 5 digits can be arranged in \( 5! \) ways. - If 0 is selected, the first digit cannot be 0, so we must select the first digit from the remaining 4 digits (1, 2, 3, 4, or 5) and arrange the other 4 digits. Step 4: Calculating the number of arrangements - If 0 is not selected, we have \( 5! = 120 \) ways. - If 0 is selected, the number of valid arrangements is \( 4 \times 4! = 4 \times 24 = 96 \). Thus, the total number of arrangements is: \[ 120 + 96 = 216. \] Thus, the total number of ways to form a five-digit number divisible by 3 is \( 216 \).

Answer 2

Problem: Form a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition. Find the total number of such numbers.

Step 1: Understand the conditions - Digits available: 0, 1, 2, 3, 4, 5 - Number length: 5 digits (no repetition) - The number must be divisible by 3. - The first digit cannot be 0 (to make it a five-digit number).

Step 2: Divisibility rule for 3 A number is divisible by 3 if the sum of its digits is divisible by 3.

Step 3: Find all 5-digit combinations of digits (without repetition) whose digit sum is divisible by 3 - Sum of all digits (0 + 1 + 2 + 3 + 4 + 5) = 15, which is divisible by 3. - We need to select 5 digits out of 6 such that their sum is divisible by 3. Possible digits to omit and their sums: - Omit 0 → sum = 15 - 0 = 15 (divisible by 3) - Omit 3 → sum = 15 - 3 = 12 (divisible by 3) - Omit other digits → sum will not be divisible by 3 Thus, the 5-digit sets whose sum is divisible by 3 are: - Set 1: {1, 2, 3, 4, 5} (omit 0) - Set 2: {0, 1, 2, 4, 5} (omit 3)

Step 4: Count the numbers formed from each set - For Set 1 (omit 0): digits are 1, 2, 3, 4, 5 Number of 5-digit numbers without repetition: \(5! = 120\) (no zero, so no restriction on leading digit) - For Set 2 (omit 3): digits are 0, 1, 2, 4, 5 First digit cannot be 0 → choices for the first digit = 4 (1, 2, 4, 5) Remaining 4 digits can be arranged in \(4! = 24\) ways Total numbers = \(4 \times 24 = 96\)

Step 5: Total numbers divisible by 3 \[ 120 + 96 = 216. \]

Final answer: \[ \boxed{216}. \]