# How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?

Mar 16, 2016

$\left\{1 , 3 , 5\right\}$

#### Explanation:

Any odd number may be represented as $2 k + 1$ for some integer $k$. Then, let's represent our odd integers as $2 k + 1 , 2 k + 3 , 2 k + 5$ where $k$ is a non-negative integer. From our initial condition, we have that

$\left(2 k + 1\right) \left(2 k + 3\right) + 24 = 3 \left(\left(2 k + 1\right) + \left(2 k + 3\right) + \left(2 k + 5\right)\right)$

Simplifying both sides, we get

$4 {k}^{2} + 8 k + 27 = 18 k + 27$

$\implies 4 {k}^{2} - 10 k = 0$

$\implies 4 k \left(k - \frac{5}{2}\right) = 0$

$\implies k = 0$ or $k = \frac{5}{2}$

But, as we specified that $k$ must be an integer, we know that $k \ne \frac{5}{2}$. Thus we have $k = 0$ and so our odd integers are $\left\{1 , 3 , 5\right\}$

Checking this, we find that $1 \cdot 3 + 24 = 27 = 3 \left(1 + 3 + 5\right)$