How do you find three consecutive odd integers such that twice the sum of the three numbers is 33 or more than the third number?

Mar 21, 2018

$5 , 7 , 9$

Explanation:

Let the three consecutive odd numbers be: $\left(2 n - 1\right) , \left(2 n + 1\right) , \left(2 n + 3\right)$ for some $n \in \mathbb{N}$

We are told that twice the sum of these numbers is at least equal to $33 + \left(2 n + 3\right)$

Thus, considering the limiting case, $2 \left(2 n - 1 + 2 n + 1 + 2 n + 3\right) = \left(2 n + 3\right) + 33$

$2 \left(6 n + 3\right) = \left(2 n + 3\right) + 33$

$12 n + 6 = 2 n + 36$

$10 n = 30$

$n = 3$

Replacing $n = 3$ into our sequence of odd numbers $\to \left\{5 , 7 , 9\right\}$

To check: $5 + 7 + 9 = 21$

$2 \times 21 = 42$

$42 - 9 = 33$