The sum of the squares of three consecutive odd numbers is 2531.

Find the smallest of these numbers.

Question

The sum of the squares of three consecutive odd numbers is 2531.

Find the smallest of these numbers.

Accepted Answer

Let the three consecutive odd numbers be $$ x-2 $$, $$ x $$, and $$ x+2 $$.  
The equation based on the sum of their squares is:

$$
(x-2)^2 + x^2 + (x+2)^2 = 2531
$$

Expanding each term:  
$$
(x-2)^2 = x^2 - 4x + 4, \quad x^2, \quad (x+2)^2 = x^2 + 4x + 4
$$

Adding these together:  
$$
x^2 - 4x + 4 + x^2 + x^2 + 4x + 4 = 2531
$$

Simplify:  
$$
3x^2 + 8 = 2531
$$

Solve for $$ x^2 $$:  
$$
3x^2 = 2531 - 8 = 2523
$$

$$
x^2 = \frac{2523}{3} = 841
$$

Take the square root:  
$$
x = \sqrt{841} = 29
$$

Thus, the three consecutive odd numbers are:  
$$ x-2 = 27, \, x = 29, \, x+2 = 31 $$.

---

### Final Answer:
The smallest number is **27**.

Correct option: **(a)**

Answer

Answer

Answer