The problem states that the sum of four consecutive odd integers is 320. We are asked to find the least of these integers.

1. Problem Description

The problem states that the sum of four consecutive odd integers is

3

2

0. We are asked to find the least of these integers.

2. Solution Steps

Let $x$ be the smallest odd integer. Then the next three consecutive odd integers are $x+2$, $x+4$, and $x+6$.

The sum of these four integers is 320, so we have the equation:

$x + (x+2) + (x+4) + (x+6) = 320$

Combine the terms on the left side:

$4x + 12 = 320$

Subtract 12 from both sides:

$4x = 320 - 12$

$4x = 308$

Divide both sides by 4:

$x = \frac{308}{4}$

$x = 77$

Therefore, the smallest odd integer is

7

7.

3. Final Answer

77