If twice the sum of three consecutive numbers is 12, and the two lowest numbers add up to 3, what is the highest number?

Options

A. `2`
B. `3`
C. `6`
D. `9`

Answer

To solve the problem step by step, we will define the three consecutive numbers and set up equations based on the information provided.

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

### Step 2: Set up the first equation
According to the problem, twice the sum of these three consecutive numbers is 12. We can express this mathematically:
\[ 2 \times (x + (x + 1) + (x + 2)) = 12 \]

### Step 3: Simplify the equation
Now, simplify the left side:
\[ 2 \times (x + x + 1 + x + 2) = 12 \]
\[ 2 \times (3x + 3) = 12 \]

### Step 4: Divide both sides by 2
Dividing both sides by 2 gives:
\[ 3x + 3 = 6 \]

### Step 5: Solve for \( x \)
Now, subtract 3 from both sides:
\[ 3x = 6 - 3 \]
\[ 3x = 3 \]
Now, divide by 3:
\[ x = 1 \]

### Step 6: Find the three consecutive numbers
Now that we have \( x \), we can find the three consecutive numbers:
- First number: \( x = 1 \)
- Second number: \( x + 1 = 1 + 1 = 2 \)
- Third number: \( x + 2 = 1 + 2 = 3 \)

### Step 7: Identify the highest number
The highest number among the three consecutive numbers \( 1, 2, 3 \) is:
\[ \text{Highest number} = 3 \]

### Final Answer
Thus, the highest number is **3**.

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