The sum of three consecutive natural numbers divisible by 3 is 45. The smallest number is :

Options

A. 18
B. 3
C. 12
D. 9

Answer

To find the smallest of three consecutive natural numbers that sum up to 45 and are divisible by 3, we can follow these steps:

### Step 1: Define the three consecutive natural numbers
Let the three consecutive natural numbers be represented as:
- First number: \( x \)
- Second number: \( x + 1 \)
- Third number: \( x + 2 \)

### Step 2: Set up the equation for their sum
The sum of these three numbers can be expressed as:
\[ x + (x + 1) + (x + 2) = 45 \]

### Step 3: Simplify the equation
Combine like terms:
\[ 3x + 3 = 45 \]

### Step 4: Isolate \( x \)
Subtract 3 from both sides:
\[ 3x = 45 - 3 \]
\[ 3x = 42 \]

### Step 5: Solve for \( x \)
Divide both sides by 3:
\[ x = \frac{42}{3} = 14 \]

### Step 6: Identify the three consecutive numbers
Now, we can find the three consecutive numbers:
- First number: \( x = 14 \)
- Second number: \( x + 1 = 15 \)
- Third number: \( x + 2 = 16 \)

### Step 7: Find the smallest number
The smallest of these three consecutive numbers is:
\[ \text{Smallest number} = 14 \]

### Final Answer
The smallest number is **14**.

---