How many four digit number are there where g.c.d. with 18 is 3

Answer

To solve the problem of finding how many four-digit numbers have a greatest common divisor (g.c.d.) of 3 with 18, we will follow these steps:

### Step 1: Understand the conditions
The g.c.d. of a number \( n \) with 18 is given as 3. Since \( 18 = 2 \times 3^2 \), for the g.c.d. to be 3, \( n \) must be a multiple of 3 but not a multiple of 9 (to avoid the g.c.d. being 9) and also not a multiple of 2 (to avoid the g.c.d. being 6).

### Step 2: Identify the range of four-digit numbers
Four-digit numbers range from 1000 to 9999.

### Step 3: Find the first and last four-digit odd multiples of 3
- The smallest four-digit number is 1000. The first odd multiple of 3 greater than or equal to 1000 is 1005.
- The largest four-digit number is 9999, which is also an odd multiple of 3.

### Step 4: Formulate the arithmetic progression (AP)
The sequence of odd multiples of 3 from 1005 to 9999 can be expressed as an arithmetic progression:
- First term \( a = 1005 \)
- Common difference \( d = 6 \) (since the next odd multiple of 3 is 3 more than the previous odd multiple of 3, and we skip the even multiples)
- Last term \( l = 9999 \)

### Step 5: Find the number of terms in the AP
Using the formula for the \( n \)-th term of an AP:
\[ l = a + (n - 1) \cdot d \]
Substituting the known values:
\[ 9999 = 1005 + (n - 1) \cdot 6 \]
Rearranging gives:
\[ 9999 - 1005 = (n - 1) \cdot 6 \]
\[ 8994 = (n - 1) \cdot 6 \]
\[ n - 1 = \frac{8994}{6} = 1499 \]
Thus, \( n = 1500 \).

### Step 6: Find the odd multiples of 9 within this range
Next, we need to find the odd multiples of 9 that fall within our range:
- The first odd multiple of 9 greater than or equal to 1000 is 1005 (which is also a multiple of 3).
- The last odd multiple of 9 less than or equal to 9999 can be found similarly.

The sequence of odd multiples of 9 can be expressed as:
- First term \( a = 1005 \)
- Common difference \( d = 18 \)

Using the same formula for the \( n \)-th term:
\[ l = a + (n - 1) \cdot d \]
Substituting the known values:
\[ 9999 = 1005 + (n - 1) \cdot 18 \]
Rearranging gives:
\[ 9999 - 1005 = (n - 1) \cdot 18 \]
\[ 8994 = (n - 1) \cdot 18 \]
\[ n - 1 = \frac{8994}{18} = 499 \]
Thus, \( n = 500 \).

### Step 7: Calculate the required count
The total number of four-digit numbers that are odd multiples of 3 is 1500. The number of those that are also odd multiples of 9 is 500.

Thus, the required count of four-digit numbers where the g.c.d. with 18 is 3:
\[ 1500 - 500 = 1000 \]

### Final Answer
The total number of four-digit numbers where the g.c.d. with 18 is 3 is **1000**.