The total number of 4-digit number whose greatest common divisor with 18 is 3, is `"______"`

Answer

To solve the problem of finding the total number of 4-digit numbers whose greatest common divisor (GCD) with 18 is 3, we can follow these steps:

### Step 1: Understand the GCD condition
The GCD of a number with 18 being 3 means that the number must be divisible by 3 but not by 6 (since 6 is the product of 2 and 3, and we want to avoid any factor of 2). Therefore, the number must be of the form:
\[ n = 3k \]
where \( k \) is an odd integer (to ensure that \( n \) is not divisible by 2).

### Step 2: Determine the range of 4-digit numbers
The smallest 4-digit number is 1000 and the largest is 9999. Therefore, we need to find the values of \( k \) such that:
\[ 1000 \leq 3k \leq 9999 \]

### Step 3: Solve for \( k \)
Dividing the entire inequality by 3 gives:
\[ \frac{1000}{3} \leq k \leq \frac{9999}{3} \]
Calculating the bounds:
\[ 333.33 \leq k \leq 3333 \]
Since \( k \) must be an integer, we round up the lower bound and round down the upper bound:
\[ 334 \leq k \leq 3333 \]

### Step 4: Count the odd integers in the range
Now we need to find the odd integers between 334 and 3333. The smallest odd integer greater than or equal to 334 is 335, and the largest odd integer less than or equal to 3333 is 3333.

The sequence of odd integers can be expressed as:
\[ 335, 337, 339, \ldots, 3333 \]

This is an arithmetic sequence where:
- First term \( a = 335 \)
- Common difference \( d = 2 \)

To find the number of terms \( n \) in this sequence, we can use the formula for the \( n \)-th term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
Setting \( a_n = 3333 \):
\[ 3333 = 335 + (n-1) \cdot 2 \]
\[ 3333 - 335 = (n-1) \cdot 2 \]
\[ 2998 = (n-1) \cdot 2 \]
\[ n-1 = 1499 \]
\[ n = 1500 \]

### Conclusion
Thus, the total number of 4-digit numbers whose GCD with 18 is 3 is **1500**.