If the square of sum of three positive consecutive natural numbers exceeds the sum of their squares by 292, then what is the largest of the three numbers?

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, let's denote the three consecutive natural numbers as \( x \), \( x + 1 \), and \( x + 2 \).

### Step 1: Write the equation for the square of the sum of the three numbers.
The sum of the three consecutive numbers is:
\[ x + (x + 1) + (x + 2) = 3x + 3 \]
Now, we square this sum:
\[ (3x + 3)^2 = 9x^2 + 18x + 9 \]

### Step 2: Write the equation for the sum of their squares.
Next, we find the sum of the squares of the three numbers:
\[ x^2 + (x + 1)^2 + (x + 2)^2 \]
Expanding this, we get:
\[ x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 3x^2 + 6x + 5 \]

### Step 3: Set up the equation based on the problem statement.
According to the problem, the square of the sum exceeds the sum of their squares by 292:
\[ (3x + 3)^2 - (x^2 + (x + 1)^2 + (x + 2)^2) = 292 \]
Substituting the expressions we derived:
\[ (9x^2 + 18x + 9) - (3x^2 + 6x + 5) = 292 \]

### Step 4: Simplify the equation.
Now, simplifying the left side:
\[ 9x^2 + 18x + 9 - 3x^2 - 6x - 5 = 292 \]
This simplifies to:
\[ 6x^2 + 12x + 4 = 292 \]

### Step 5: Rearrange the equation.
Now, we rearrange it to set it to zero:
\[ 6x^2 + 12x + 4 - 292 = 0 \]
\[ 6x^2 + 12x - 288 = 0 \]

### Step 6: Simplify the equation further.
Dividing the entire equation by 6 to simplify:
\[ x^2 + 2x - 48 = 0 \]

### Step 7: Factor the quadratic equation.
Now we need to factor this quadratic equation:
\[ (x + 8)(x - 6) = 0 \]

### Step 8: Solve for \( x \).
Setting each factor to zero gives us:
\[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \quad (\text{not a natural number}) \]
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \quad (\text{valid solution}) \]

### Step 9: Determine the largest number.
The three consecutive natural numbers are:
\[ 6, 7, 8 \]
Thus, the largest of these three numbers is:
\[ \boxed{8} \]