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

27

We have,

The sum of the squares of 3 consecutive odd numbers is 2531.

According to the question,

⇒ (2n - 1)² + (2n + 1)² + (2n + 3)²  = 2531

Simplifying the equation:

⇒ 4n² - 4n + 1 + 4n² + 4n + 1 + 4n² + 12n + 9 = 2531

⇒ 12n² + 12n + 11 = 2531

⇒ 12n² + 12n - 2520 = 0

Dividing both sides of the equation by 12:

⇒ n² + n - 210 = 0

Now, we can factorize the quadratic equation:

⇒ (n + 15)(n - 14) = 0

So, we have two possible solutions for n: n = -15 and n = 14.

However, n represents an integer, so the valid solution is n = 14.

Now, we can find the three consecutive odd numbers:

First odd number: 2n - 1 = 2(14) - 1 = 27

Second odd number: 2n + 1 = 2(14) + 1 = 29

Third odd number: 2n + 3 = 2(14) + 3 = 31

So  the  smallest odd number is = 27