The total number of integers between 100 and 200, which are divisible by both 9 and 6. is :

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem of finding the total number of integers between 100 and 200 that are divisible by both 9 and 6, we can follow these steps:

### Step 1: Find the LCM of 9 and 6
To determine the integers that are divisible by both 9 and 6, we first need to find the least common multiple (LCM) of these two numbers.

- The prime factorization of 9 is \(3^2\).
- The prime factorization of 6 is \(2^1 \times 3^1\).

The LCM is found by taking the highest power of each prime factor:
- For 2: \(2^1\)
- For 3: \(3^2\)

Thus, the LCM of 9 and 6 is:
\[ LCM(9, 6) = 2^1 \times 3^2 = 2 \times 9 = 18 \]

### Step 2: Identify the range
We need to find integers between 100 and 200 that are divisible by 18.

### Step 3: Find the smallest multiple of 18 greater than 100
To find the smallest integer greater than 100 that is divisible by 18, we can divide 100 by 18 and round up to the nearest whole number:
\[ \frac{100}{18} \approx 5.56 \]
Rounding up gives us 6. Now, we calculate:
\[ 18 \times 6 = 108 \]

### Step 4: Find the largest multiple of 18 less than 200
Next, we find the largest integer less than 200 that is divisible by 18:
\[ \frac{200}{18} \approx 11.11 \]
Rounding down gives us 11. Now, we calculate:
\[ 18 \times 11 = 198 \]

### Step 5: List the multiples of 18 between 100 and 200
The multiples of 18 between 100 and 200 are:
- 108 (18 × 6)
- 126 (18 × 7)
- 144 (18 × 8)
- 162 (18 × 9)
- 180 (18 × 10)
- 198 (18 × 11)

### Step 6: Count the multiples
Now we count the multiples listed:
- 108
- 126
- 144
- 162
- 180
- 198

There are a total of 6 integers between 100 and 200 that are divisible by both 9 and 6.

### Final Answer
The total number of integers between 100 and 200 that are divisible by both 9 and 6 is **6**.

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