# Question #cb56b

May 19, 2017

$\left\{2 n + 1 , 2 n + 3 , 2 n + 5\right\} = \left\{11 , 13 , 15\right\}$

#### Explanation:

If you have three, consecutive odd integers, you could write it like this
$\left\{2 n + 1 , 2 n + 3 , 2 n + 5\right\}$ where $n$ is an integer

The $2 n$ term is guaranteed to be even. So adding an odd value (i.e., 1, 3, and 5) ensures it will be odd.

The smallest odd is $2 n + 1$. The largest is $2 n + 5$. The problem tells us that

$2 \left(2 n + 1\right) = 2 n + 5 + 7$
$4 n + 2 = 2 n + 12$
$2 n + 2 = 12$ (subtract $2 n$ from both sides)
$2 n = 10$ (subtract $2$ from both sides)
$n = 5$

Plugging $n = 5$ back into our three consecutive integers gives
$\left\{2 n + 1 , 2 n + 3 , 2 n + 5\right\} = \left\{11 , 13 , 15\right\}$

May 19, 2017

Let $x$ be the smallest of the three consecutive odd integers. Then the three consecutive odd integers are:
$x , x + 2 , x + 4$

$2 x = 7 + \left(x + 4\right)$
Left side of the equation comes from "Twice the smallest"
Right side of the equation comes from "seven more than the largest"

$2 x = x + 11$

x=11

Since we have $x$ now, the three consecutive odd integers are:
$11 , 11 + 2 , 11 + 4$

$11 , 13 , 15$