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

1 answer

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:

\[ \boxed{-5} \]