The number of positive six-digit integers which are divisible by `9` and four of its digits are `1`, `0`, `0`, `5` is

Options

A. `60`
B. `120`
C. `180`
D. `210`

Explanation

To solve the problem of finding the number of positive six-digit integers that are divisible by 9 and contain the digits 1, 0, 0, and 5, we will follow these steps:

### Step 1: Determine the required sum of digits for divisibility by 9
To check for divisibility by 9, the sum of the digits of the number must be divisible by 9. The digits we have are 1, 0, 0, and 5.

Calculating the initial sum:
\[ 1 + 0 + 0 + 5 = 6 \]

To make the total sum divisible by 9, we need to add digits such that the total becomes 9 or 18. Since we are forming a six-digit number, we can only reach 9. Therefore, we need to add digits that sum to:
\[ 9 - 6 = 3 \]

### Step 2: Identify possible pairs of digits that sum to 3
We can find pairs of digits that can be added to our existing digits (1, 0, 0, 5) to achieve a total sum of 9. The possible pairs of digits that sum to 3 are:
1. (0, 3)
2. (1, 2)

### Step 3: Case 1 - Using the digits (0, 3)
In this case, our digits are 1, 0, 0, 5, 0, and 3.

#### Step 3.1: Count the arrangements
We need to arrange the digits 1, 0, 0, 5, 0, and 3. However, we cannot have 0 in the leading position since we are forming a six-digit number.

The total arrangements of the digits without restriction is given by:
\[ \frac{6!}{3!} = \frac{720}{6} = 120 \]
(where 3! accounts for the three 0s).

#### Step 3.2: Exclude invalid arrangements
Now, we need to subtract the cases where 0 is in the leading position. If 0 is the first digit, we are left with the digits 0, 1, 5, and 3 to arrange, which can be arranged as:
\[ \frac{5!}{2!} = \frac{120}{2} = 60 \]
(where 2! accounts for the two 0s).

Thus, the valid arrangements for this case are:
\[ 120 - 60 = 60 \]

### Step 4: Case 2 - Using the digits (1, 2)
In this case, our digits are 1, 0, 0, 5, 1, and 2.

#### Step 4.1: Count the arrangements
The total arrangements of the digits 1, 0, 0, 5, 1, and 2 is given by:
\[ \frac{6!}{2! \cdot 2!} = \frac{720}{4} = 180 \]
(where 2! accounts for the two 1s and 2! accounts for the two 0s).

#### Step 4.2: Exclude invalid arrangements
Again, we need to exclude cases where 0 is in the leading position. If 0 is the first digit, we are left with the digits 1, 0, 5, 1, and 2 to arrange:
\[ \frac{5!}{2!} = \frac{120}{2} = 60 \]

Thus, the valid arrangements for this case are:
\[ 180 - 60 = 120 \]

### Step 5: Total valid arrangements
Now, we add the valid arrangements from both cases:
\[ 60 \text{ (from case 1)} + 120 \text{ (from case 2)} = 180 \]

### Final Answer
The total number of positive six-digit integers which are divisible by 9 and contain the digits 1, 0, 0, and 5 is **180**.

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