# How do you find three consecutive odd integers such that the sum of all three is 36 less than the product of the smaller two?

Dec 11, 2016

$7 , 9 , 11$.

#### Explanation:

Let the numbers be $x$, $x + 2$ and $x + 4$

$x + \left(x + 2\right) + \left(x + 4\right) = x \left(x + 2\right) - 36$

$x + x + 2 + x + 4 = {x}^{2} + 2 x - 36$

$3 x + 6 = {x}^{2} + 2 x - 36$

$0 = {x}^{2} - x - 42$

$0 = \left(x - 7\right) \left(x + 6\right)$

$x = 7 \mathmr{and} - 6$

Since the numbers have to be odd, the numbers are $7$, $9$ and $11$.

Hopefully this helps!