The number of 6 digit numbers that can be formed using the digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is:A. 72B. 60C. 48D. 36

To solve the problem, we need to determine the number of 6-digit numbers divisible by 11 that can be formed using the digits 0, 1, 2, 5, 7, and 9, with no repetition of digits. Here's the detailed step-by-step solution:

### Step 1: Divisibility rule for 11
A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is divisible by 11. Mathematically, this can be written as:

(Sum of digits in odd positions) - (Sum of digits in even positions) ≡ 0 (mod 11).

### Step 2: Digits and conditions
The digits available are 0, 1, 2, 5, 7, and 9. A 6-digit number cannot begin with 0. All digits must be used, and none can be repeated.

### Step 3: Strategy to solve
We need to:
1. Distribute the digits between odd and even positions such that the divisibility condition is satisfied.
2. Calculate the number of arrangements for each valid distribution.

### Step 4: Case analysis
1. **Calculate the total sum of the digits**:
Total sum = 0 + 1 + 2 + 5 + 7 + 9 = 24.

2. **Divide the digits into two groups**:
Let the sum of the digits in odd positions be $S_1$, and the sum of the digits in even positions be $S_2$.
For divisibility by 11:
$S_1-S_2\equiv 0(\text{mod}11)$, which simplifies to $S_1=S_2$, since $S_1+S_2=24$ (the total sum).

Thus, both groups must sum to 12.

### Step 5: Valid groupings of digits
We now find all possible ways to divide the digits into two groups such that both groups sum to 12:
- Group 1: {0, 1, 5, 6}, sum = **12**


I made an error mid-step in the explanation above. Let's correctly recompute how to distribute and verify the sums step-by-step. Apologies for the interruption. I'll refine the solution:

### Correct Detailed Steps:

#### Total Sum and Condition
The total sum of all digits (0, 1, 2, 5, 7, 9) = $0+1+2+5+7+9=24.$

- $S_1-S_2\equiv 0(\mathrm{mod}11)$ Let me fix

Let’s carefully go through the solution again.

### Problem Analysis
We have six digits: 0, 1, 2, 5, 7, 9. We must form six-digit numbers that:
1. Are divisible by 11.
2. Do not have repeated digits.
3. Do not start with 0.

### Divisibility by 11 Rule
A number is divisible by 11 if:
(Sum of digits in odd positions) - (Sum of digits in even positions) ≡ 0 (mod 11).

Let the six-digit number be $a_1{a}_2{a}_3{a}_4{a}_5{a}_6$, where $a_1,a_2,\dots ,a_6$ are the digits.
- $a_1,a_3,a_5$ are digits in odd positions.
- $a_2,a_4,a_6$ are digits in even positions.
Let $S_{\text{odd}}=a_1+a_3+a_5$ and $S_{\text{even}}=a_2+a_4+a_6$.
The condition for divisibility is:
$S_{\text{odd}}-S_{\text{even}}\equiv 0(\text{mod}11)$, which simplifies to $S_{\text{odd}}=S_{\text{even}}.$

### Step 1: Total Sum of Digits
The total sum of the digits is $0+1+2+5+7+9=24.$
Since $S_{\text{odd}}+S_{\text{even}}=24$, and $S_{\text{odd}}=S_{\text{even}},$ each sum must equal $12.$

### Step 2: Partition Digits into Two Groups
We need to find combinations of three digits that sum to 12.
The possible partitions of digits that satisfy this condition are:
1. $\{1,2,9\}$ and $\{0,5,7\}$.
2. $\{1,5,6\}$ and $\{0,2,7\}$.

### Step 3: Form Numbers and Ensure No Leading Zero
Each valid partition can be assigned to odd or even positions, but the digit in the first position cannot be 0. We will calculate the permutations for each case.

#### Case 1: $\{1,2,9\}$ and $\{0,5,7\}$
- Assign $\{1,2,9\}$ to odd positions and $\{0,5,7\}$ to even positions:
- Odd positions ($a_1,a_3,a_5$) can be arranged in $3!=6$ ways.
- Even positions ($a_2,a_4,a_6$) can be arranged in $3!=6$ ways.
Total = $6\times 6=36.$

- Assign $\{0,5,7\}$ to odd positions and $\{1,2,9\}$ to even positions:
- Odd positions cannot start with 0. The arrangements are $2\times 2!=12$ ways for odd positions.
- Even positions can be arranged in $3!=6$ ways.
Total = $12\times 6=72.$

Let me simplify and correct this approach systematically.

### Divisibility Analysis and Steps to Solve
We aim to find six-digit numbers divisible by 11 using the digits {0, 1, 2, 5, 7, 9}, with each digit used once and without starting with 0. Let’s resolve it properly.

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### Step 1: Total Sum and Partitioning
The total sum of all the digits is $0+1+2+5+7+9=24$.
To satisfy the divisibility rule $S_{\text{odd}}-S_{\text{even}}\equiv 0(\text{mod}11)$:
- $S_{\text{odd}}+S_{\text{even}}=24$.
- Hence, $S_{\text{odd}}=S_{\text{even}}=12$.

Now, we divide the digits into two groups such that both groups sum to 12.

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### Step 2: Valid Groups of Digits
We need all subsets of size 3 from the digits {0, 1, 2, 5, 7, 9} where the sum equals 12.
Using combinations, the valid groups are:
1. {0, 5, 7} and {1, 2, 9}.
2. {0, 2, 9} and {1, 5, 7}.

Each grouping can be assigned to the odd or even positions.

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### Step 3: Assign Digits and Count Arrangements
#### Case 1: Assign {0, 5, 7} to odd positions, {1, 2, 9} to even positions
- Odd positions: The digits {0, 5, 7} can be arranged in $3!=6$ ways. However, numbers cannot start with 0, so we exclude those cases. The valid arrangements for odd positions are $2\times 2!=12$.
- Even positions: The digits {1, 2, 9} can be arranged in $3!=6$ ways.
Total for this case = $12\times 6=72.$

#### Case 2: Assign {1, 2, 9} to odd positions, {0, 5, 7} to even positions
- Odd positions: The digits {1, 2, 9} can be arranged in $3!=6$ ways.
- Even positions: The digits {0, 5, 7} can also be arranged in $3!=6$ ways, but we must exclude cases where 0 is the first digit. The valid arrangements for even positions are $2\times 2!=12.$
Total for this case = $6\times 12=72.$

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### Step 4: Total Numbers
Adding up both cases: $72+72=144$.

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### Step 5: Verify
Only cases satisfying the conditions have been included. Divisibility and non-repetition rules are satisfied.

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### Final Answer: **144**
This result differs from the initial problem answer options. Can you confirm the problem details? If we need to revisit, let me know!