# If the sum of three consecutive odd integers is 129, what are the integers?

Feb 3, 2016

$41 , 43 , \mathmr{and} 45$

#### Explanation:

Let the smallest of the three consecutive odd numbers be $n$

Then the three odd numbers will be: $n , n + 2 , n + 4$

We are told
$\textcolor{w h i t e}{\text{XXX}} n + \left(n + 2\right) + \left(n + 4\right) = 129$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 3 n + 6 = 129$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} 3 n = 123$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} n = 41$

and the three consecutive odd numbers are $41 , \left(41 + 2\right) , \left(41 + 4\right)$

Feb 3, 2016

Let the first number be $\omega$
The difference of consecutive odd numbers is $2$
So,the next two numbers are $\omega + 2$ , $\omega + 4$

Then,

$\rightarrow \left(\omega\right) + \left(\omega + 2\right) + \left(\omega + 4\right) = 129$

Remove brackets:

$\rightarrow \omega + \omega + 2 + \omega + 4 = 129$

$\rightarrow 3 \omega + 6 = 129$

$\rightarrow 3 \omega = 129 - 6$

$\rightarrow 3 \omega = 123$

$\rightarrow \omega = \frac{123}{3} = 41$

So,$\omega + 2 = 41 + 2 = 43 , \omega + 4 = 41 + 4 = 45$

The three integers are $41 , 43 \mathmr{and} 45$