# Four consecutive odd integers add up to 64. What are the numbers?

Apr 26, 2016

$13 , 15 , 17 \mathmr{and} 19$

#### Explanation:

Let the first odd number be $= 2 n + 1$, where $n$ is any positive integer.

Thus we have four consecutive odd numbers
$\left(2 n + 1\right) , \left(2 n + 3\right) , \left(2 n + 5\right) \mathmr{and} \left(2 n + 7\right)$
Setting the sum of these numbers equal to the given value

$\left(2 n + 1\right) + \left(2 n + 3\right) + \left(2 n + 5\right) + \left(2 n + 7\right) = 64$, simplifying
$\left(8 n + 16\right) = 64$, dividing both sides and solving for $n$
$\left(n + 2\right) = 8$
or $n = 8 - 2 = 6$
The numbers are
$\left(2 \times 6 + 1\right) , \left(2 \times 6 + 3\right) , \left(2 \times 6 + 5\right) \mathmr{and} \left(2 \times 6 + 7\right)$
$13 , 15 , 17 \mathmr{and} 19$