How many numbers are co-prime to 4608 that lie between 1000 and 2000 ?

Options

A. 249
B. 332
C. 166
D. 334

Answer

To find how many numbers are co-prime to 4608 that lie between 1000 and 2000, we can follow these steps:

### Step 1: Factorize 4608
First, we need to factor 4608 into its prime factors.

4608 can be expressed as:
\[ 4608 = 2^9 \times 3^2 \]

### Step 2: Identify co-prime conditions
A number is co-prime to 4608 if it is not divisible by the prime factors of 4608, which are 2 and 3. Therefore, we need to find numbers between 1000 and 2000 that are not multiples of 2 or 3.

### Step 3: Count the total numbers between 1000 and 2000
The total numbers from 1000 to 2000 (inclusive) can be calculated as:
\[ 2000 - 1000 + 1 = 1001 \]

### Step 4: Count the even numbers (multiples of 2)
Even numbers between 1000 and 2000 can be calculated as follows:
- The first even number is 1000 and the last is 2000.
- The sequence of even numbers is 1000, 1002, ..., 2000.

This forms an arithmetic sequence where:
- First term (a) = 1000
- Common difference (d) = 2
- Last term (l) = 2000

To find the number of terms (n):
\[ n = \frac{l - a}{d} + 1 = \frac{2000 - 1000}{2} + 1 = 501 \]

### Step 5: Count the multiples of 3
Next, we need to count how many numbers between 1000 and 2000 are multiples of 3.
- The first multiple of 3 in this range is 1002 and the last is 1998.
- The sequence of multiples of 3 is 1002, 1005, ..., 1998.

This also forms an arithmetic sequence where:
- First term (a) = 1002
- Common difference (d) = 3
- Last term (l) = 1998

To find the number of terms (n):
\[ n = \frac{l - a}{d} + 1 = \frac{1998 - 1002}{3} + 1 = 333 \]

### Step 6: Count the multiples of 6 (common multiples of 2 and 3)
Now, we need to count how many numbers are multiples of both 2 and 3 (i.e., multiples of 6).
- The first multiple of 6 in this range is 1002 and the last is 1998.
- The sequence of multiples of 6 is 1002, 1008, ..., 1998.

This forms an arithmetic sequence where:
- First term (a) = 1002
- Common difference (d) = 6
- Last term (l) = 1998

To find the number of terms (n):
\[ n = \frac{l - a}{d} + 1 = \frac{1998 - 1002}{6} + 1 = 167 \]

### Step 7: Apply the principle of inclusion-exclusion
To find the total numbers that are either multiples of 2 or multiples of 3, we use:
\[ \text{Multiples of 2 or 3} = \text{Multiples of 2} + \text{Multiples of 3} - \text{Multiples of 6} \]
\[ = 501 + 333 - 167 = 667 \]

### Step 8: Calculate co-prime numbers
Finally, to find the numbers that are co-prime to 4608, we subtract the count of numbers that are multiples of 2 or 3 from the total count of numbers:
\[ \text{Co-prime numbers} = \text{Total numbers} - \text{Multiples of 2 or 3} \]
\[ = 1001 - 667 = 334 \]

### Conclusion
Thus, the number of integers between 1000 and 2000 that are co-prime to 4608 is **334**.