# How do you find all sets of three consecutive odd integers whose sum is between 20 and 30?

Feb 11, 2016

Write the odd numbers in the form $2 k + 1 , 2 k + 3 , 2 k + 5$ and set up an inequality to find all valid values for $k$ to find that the sets are
$\left\{5 , 7 , 9\right\}$ and $\left\{7 , 9 , 11\right\}$

#### Explanation:

Any set of three consecutive odd integers may be written as
$\left\{2 k + 1 , 2 k + 3 , 2 k + 5\right\}$ for some $k \in \mathbb{Z}$

Then, we just need to find a condition on $k$ such that
$20 < \left(2 k + 1\right) + \left(2 k + 3\right) + \left(2 k + 5\right) < 30$

Simplifying, we get

$20 < 6 k + 9 < 30$

Subtracting $9$ gives us

$11 < 6 k < 21$

Dividing by $6$

$\frac{11}{6} < k < \frac{21}{6}$

Thus we will have a set with the desired property when $k$ is an integer between $\frac{11}{6}$ and $\frac{21}{6}$, that is, when $k = 2$ or $k = 3$. This gives us the result that the only such sets are

$\left\{2 \left(2\right) + 1 , 2 \left(2\right) + 3 , 2 \left(2\right) + 5\right\} = \left\{5 , 7 , 9\right\}$
and
$\left\{2 \left(3\right) + 1 , 2 \left(3\right) + 3 , 2 \left(3\right) + 5\right\} = \left\{7 , 9 , 11\right\}$