Understanding the Problem of Forming 6-Digit Numbers

The question asks us to find the total count of 6-digit numbers that can be created using each of the digits 0, 1, 2, 3, 4, and 5 exactly once. Additionally, these numbers must be divisible by 6.

This problem involves permutations and the application of divisibility rules.

Divisibility Rules for 6

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

• Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). In our set of available digits {0, 1, 2, 3, 4, 5}, the even digits are 0, 2, and 4.
• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Let's find the sum of the given digits: $0 + 1 + 2 + 3 + 4 + 5 = 15$. Since 15 is divisible by 3 ($15 \div 3 = 5$), any number formed by using these six digits exactly once will automatically be divisible by 3.

Therefore, the condition for the 6-digit number to be divisible by 6 simplifies to just one condition: the number must be even.

Constraints on the 6-Digit Number

A 6-digit number $d_1 d_2 d_3 d_4 d_5 d_6$ formed using the digits {0, 1, 2, 3, 4, 5} exactly once must satisfy:

• The first digit ($d_1$) cannot be 0 (as it's a 6-digit number).
• Each digit from the set {0, 1, 2, 3, 4, 5} must be used exactly once.
• The last digit ($d_6$) must be even, i.e., $d_6 \in \{0, 2, 4\}$.

Step-by-Step Solution: Counting Even 6-Digit Numbers

We will count the numbers by considering the possible even digits for the last position ($d_6$).

Case 1: The last digit is 0 ($d_6 = 0$)

If the last digit is 0, the remaining digits to be arranged in the first five positions ($d_1 d_2 d_3 d_4 d_5$) are {1, 2, 3, 4, 5}.

• The first digit ($d_1$) can be any of the 5 remaining digits (no restriction as 0 is already used at the end).
• The remaining 4 digits ($d_2, d_3, d_4, d_5$) can be arranged in $4!$ ways.

The number of ways to arrange the remaining 5 digits in the first 5 places is $5!$.

Number of numbers ending in 0 = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Total for Case 1: $5! = 120$.

Case 2: The last digit is 2 ($d_6 = 2$)

If the last digit is 2, the remaining digits to be arranged in the first five positions ($d_1 d_2 d_3 d_4 d_5$) are {0, 1, 3, 4, 5}.

• The first digit ($d_1$) cannot be 0. It can be any of {1, 3, 4, 5} (4 options).
• The remaining 4 digits ($d_2, d_3, d_4, d_5$) must be arranged in the remaining 4 positions. These 4 digits are {0} and the 3 digits from {1, 3, 4, 5} that were not chosen for $d_1$. The number of ways to arrange these 4 digits is $4!$.

Number of ways:

• Choose $d_6$: 1 option (must be 2).
• Choose $d_1$: 4 options (cannot be 0 or 2).
• Arrange the remaining 4 digits in $d_2 d_3 d_4 d_5$: $4!$ ways.

Number of numbers ending in 2 = $4 \times 4! = 4 \times 24 = 96$.

Case 3: The last digit is 4 ($d_6 = 4$)

If the last digit is 4, the remaining digits to be arranged in the first five positions ($d_1 d_2 d_3 d_4 d_5$) are {0, 1, 2, 3, 5}. This case is symmetric to Case 2.

• The first digit ($d_1$) cannot be 0. It can be any of {1, 2, 3, 5} (4 options).
• The remaining 4 digits ($d_2, d_3, d_4, d_5$) must be arranged in the remaining 4 positions. These 4 digits are {0} and the 3 digits from {1, 2, 3, 5} that were not chosen for $d_1$. The number of ways to arrange these 4 digits is $4!$.

Number of ways:

• Choose $d_6$: 1 option (must be 4).
• Choose $d_1$: 4 options (cannot be 0 or 4).
• Arrange the remaining 4 digits in $d_2 d_3 d_4 d_5$: $4!$ ways.

Number of numbers ending in 4 = $4 \times 4! = 4 \times 24 = 96$.

Total Number of Valid 6-Digit Numbers

To find the total number of 6-digit numbers that satisfy the conditions, we sum the counts from the three cases:

Total = (Numbers ending in 0) + (Numbers ending in 2) + (Numbers ending in 4)

Total = $120 + 96 + 96 = 312$.

Thus, there are 312 such 6-digit numbers.

Revision Table: Forming 6-Digit Numbers Divisible by 6

Additional Information: Permutations and Constraints

This problem is a classic example of counting permutations with restrictions. When forming numbers from a given set of digits, common restrictions include:

• The number of digits in the formed number.
• Whether digits can be repeated or must be used exactly once.
• Conditions on specific digit positions (e.g., first digit cannot be 0).
• Conditions based on divisibility rules (e.g., divisible by 2, 3, 5, 6, 10).

When solving such problems, it is often helpful to address the most restrictive conditions first, such as the non-zero first digit or the required last digit for divisibility.

Permutations of $n$ distinct objects is $n!$. The number of permutations of $n$ objects taken $r$ at a time is $P(n, r) = \frac{n!}{(n-r)!}$. When some digits are fixed or restricted, we break down the problem by considering those positions and then arranging the remaining digits in the remaining positions.