The problem is to find a set of 4 consecutive integers that sum to $-18$.

1. Problem Description

The problem is to find a set of 4 consecutive integers that sum to $-18$.

2. Solution Steps

Let the four consecutive integers be $n$, $n+1$, $n+2$, and $n+3$. Their sum is given by:

$n + (n+1) + (n+2) + (n+3) = -18$

Combining like terms, we have:

$4n + 6 = -18$

Subtract 6 from both sides:

$4n = -18 - 6$

$4n = -24$

Divide both sides by 4:

$n = \frac{-24}{4}$

$n = -6$

So the four consecutive integers are $-6$, $-6+1 = -5$, $-6+2 = -4$, and $-6+3 = -3$.

Let's check if their sum is -18:

$-6 + (-5) + (-4) + (-3) = -6 - 5 - 4 - 3 = -18$

3. Final Answer

The four consecutive integers are -6, -5, -4, -3.