Question

Four consecutive odd integers have a sum of –8. What is the value of the least integer?

Answer 1

Let the four consecutive odd integers be represented as:

\[
x, \quad x+2, \quad x+4, \quad x+6
\]

where \(x\) is the smallest of the four odd integers. According to the problem, the sum of these integers is equal to -8:

\[
x + (x + 2) + (x + 4) + (x + 6) = -8
\]

We can simplify the left-hand side:

\[
x + x + 2 + x + 4 + x + 6 = 4x + 12
\]

Thus, we have the equation:

\[
4x + 12 = -8
\]

Next, we will solve for \(x\). First, subtract 12 from both sides:

\[
4x = -8 - 12
\]

\[
4x = -20
\]

Now, divide both sides by 4:

\[
x = \frac{-20}{4} = -5
\]

The least integer, which is the first odd integer, is therefore:

\[
\boxed{-5}
\]

To verify, we'll compute the four consecutive odd integers and their sum:

The integers are:

\[
-5, \quad -3, \quad -1, \quad 1
\]

Now we calculate the sum:

\[
-5 + (-3) + (-1) + 1 = -5 - 3 - 1 + 1 = -8
\]

The sum is indeed -8, confirming that our solution is correct. Thus, the least integer is:

\[
\boxed{-5}
\]