How many 3-digit numbers are completely divisible by 6?

Options

A. 149
B. 150
C. 151
D. 166

Answer

To find how many 3-digit numbers are completely divisible by 6, we can follow these steps:

### Step 1: Identify the range of 3-digit numbers
The smallest 3-digit number is 100, and the largest is 999.

### Step 2: Find the smallest 3-digit number divisible by 6
To find the smallest 3-digit number divisible by 6, we divide 100 by 6 and check the remainder:
- \( 100 \div 6 = 16 \) remainder \( 4 \)

Since 100 is not divisible by 6, we add the difference to reach the next multiple of 6:
- \( 100 + (6 - 4) = 100 + 2 = 102 \)

So, the smallest 3-digit number divisible by 6 is **102**.

### Step 3: Find the largest 3-digit number divisible by 6
Next, we check the largest 3-digit number, which is 999:
- \( 999 \div 6 = 166 \) remainder \( 3 \)

Since 999 is not divisible by 6, we subtract the remainder to reach the previous multiple of 6:
- \( 999 - 3 = 996 \)

So, the largest 3-digit number divisible by 6 is **996**.

### Step 4: Determine the sequence of 3-digit numbers divisible by 6
Now we have a sequence of numbers starting from 102 to 996, with a common difference of 6. This forms an arithmetic sequence:
- First term \( a = 102 \)
- Last term \( l = 996 \)
- Common difference \( d = 6 \)

### Step 5: Use the formula for the nth term of an arithmetic sequence
The nth term of an arithmetic sequence can be calculated using the formula:
\[ l = a + (n - 1) \cdot d \]

Substituting the values we have:
\[ 996 = 102 + (n - 1) \cdot 6 \]

### Step 6: Solve for n
Rearranging the equation:
\[ 996 - 102 = (n - 1) \cdot 6 \]
\[ 894 = (n - 1) \cdot 6 \]
\[ n - 1 = \frac{894}{6} \]
\[ n - 1 = 149 \]
\[ n = 150 \]

### Conclusion
Thus, the total number of 3-digit numbers that are completely divisible by 6 is **150**.

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