Key Concepts and Formulas

• Divisibility Rule for 11: A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is a multiple of 11 (including 0).
• Permutations: The number of ways to arrange $n$ distinct objects in $n$ places is $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
• Basic Arithmetic: Addition, subtraction, and solving simple linear equations.

Step-by-Step Solution

Step 1: Define variables and state the divisibility rule

We are given the digits $\{0, 1, 2, 5, 7, 9\}$ and we need to form a 6-digit number divisible by 11 without repetition. Let the 6-digit number be $a_1 a_2 a_3 a_4 a_5 a_6$. According to the divisibility rule of 11, the alternating sum of the digits must be a multiple of 11. Therefore, $(a_1 + a_3 + a_5) - (a_2 + a_4 + a_6) = 11K$ where $K$ is an integer. Let $S_{odd} = a_1 + a_3 + a_5$ and $S_{even} = a_2 + a_4 + a_6$. Then, $S_{odd} - S_{even} = 11K \quad \text{(Equation 1)}$

Step 2: Establish a second equation relating the sums

The sum of all the digits is constant: $S_{odd} + S_{even} = 0 + 1 + 2 + 5 + 7 + 9 = 24 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations for $S_{odd}$ and $S_{even}$

We have a system of two linear equations:

1. $S_{odd} - S_{even} = 11K$
2. $S_{odd} + S_{even} = 24$

Adding Equation 1 and Equation 2: $2S_{odd} = 24 + 11K$ $S_{odd} = 12 + \frac{11K}{2}$

Subtracting Equation 1 from Equation 2: $2S_{even} = 24 - 11K$ $S_{even} = 12 - \frac{11K}{2}$

Since $S_{odd}$ and $S_{even}$ must be integers, $K$ must be an even integer. Let $K = 2m$, where $m$ is an integer. Substituting $K=2m$: $S_{odd} = 12 + 11m$ $S_{even} = 12 - 11m$

Step 4: Determine the possible values of m

The minimum sum of 3 digits from the set is $0 + 1 + 2 = 3$, and the maximum sum is $5 + 7 + 9 = 21$. Therefore, $3 \le S_{odd} \le 21$ and $3 \le S_{even} \le 21$.

If $m = 0$: $S_{odd} = 12$ and $S_{even} = 12$. This is a valid solution. If $m = 1$: $S_{odd} = 23$ and $S_{even} = 1$. This is not a valid solution since $S_{odd} > 21$. If $m = -1$: $S_{odd} = 1$ and $S_{even} = 23$. This is not a valid solution since $S_{odd} < 3$.

The only possible solution is $S_{odd} = 12$ and $S_{even} = 12$.

Step 5: Find the sets of digits that sum to 12

We need to find two sets of three digits from $\{0, 1, 2, 5, 7, 9\}$ such that each set sums to 12. By inspection, we find the following sets:

• $\{0, 5, 7\}$ sums to 12.
• $\{1, 2, 9\}$ sums to 12.

These are the only two sets that satisfy the condition.

Step 6: Calculate the number of 6-digit numbers when odd places are $\{0, 5, 7\}$ and even places are $\{1, 2, 9\}$

The digits in the odd places ($a_1, a_3, a_5$) are $\{0, 5, 7\}$. Since $a_1$ cannot be 0, there are 2 choices for $a_1$ (5 or 7). Then, there are 2 choices for $a_3$ and 1 choice for $a_5$. The number of arrangements for odd places is $2 \times 2 \times 1 = 4$.

The digits in the even places ($a_2, a_4, a_6$) are $\{1, 2, 9\}$. There are 3 choices for $a_2$, 2 choices for $a_4$, and 1 choice for $a_6$. The number of arrangements for even places is $3 \times 2 \times 1 = 6$.

The total number of 6-digit numbers in this case is $4 \times 6 = 24$.

Step 7: Calculate the number of 6-digit numbers when odd places are $\{1, 2, 9\}$ and even places are $\{0, 5, 7\}$

The digits in the odd places ($a_1, a_3, a_5$) are $\{1, 2, 9\}$. Since $a_1$ cannot be 0, and none of these digits are 0, there are 3 choices for $a_1$, 2 choices for $a_3$, and 1 choice for $a_5$. The number of arrangements for odd places is $3 \times 2 \times 1 = 6$.

The digits in the even places ($a_2, a_4, a_6$) are $\{0, 5, 7\}$. There are 3 choices for $a_2$, 2 choices for $a_4$, and 1 choice for $a_6$. The number of arrangements for even places is $3 \times 2 \times 1 = 6$.

The total number of 6-digit numbers in this case is $6 \times 6 = 36$.

Step 8: Calculate the total number of 6-digit numbers

The total number of 6-digit numbers divisible by 11 is the sum of the numbers from both cases: $24 + 36 = 60$.

Common Mistakes & Tips

• Forgetting the $a_1 \ne 0$ restriction: This is a common mistake that leads to overcounting. Always remember to account for this restriction when arranging digits.
• Incorrectly applying the divisibility rule: Double-check the divisibility rule for 11 and ensure that you are calculating the alternating sum correctly.
• Missing possible digit combinations: Be systematic when searching for digit combinations that sum to a specific value.

Summary

We used the divisibility rule for 11 to determine the possible sums of digits in the odd and even places. We then found the unique sets of digits that satisfied these sums and calculated the number of permutations for each case, considering the restriction that the first digit cannot be zero. The total number of 6-digit numbers divisible by 11 that can be formed using the given digits without repetition is 60.

The final answer is $\boxed{60}$, which corresponds to option (B).