Question

Four consecutive integers add up to –46. What is the least of these integers?

Answer 1

Let the four consecutive integers be represented as \( x, x+1, x+2, x+3 \).

To find their sum, we can set up the equation:

\[
x + (x+1) + (x+2) + (x+3) = -46
\]

This simplifies to:

\[
4x + 6 = -46
\]

Next, we can isolate \( x \) by subtracting 6 from both sides:

\[
4x = -46 - 6
\]
\[
4x = -52
\]

Now, divide both sides by 4 to solve for \( x \):

\[
x = \frac{-52}{4} = -13
\]

Now that we have \( x \), we can find the four consecutive integers:

- First integer: \( x = -13 \)
- Second integer: \( x + 1 = -12 \)
- Third integer: \( x + 2 = -11 \)
- Fourth integer: \( x + 3 = -10 \)

The least of these integers is \( \boxed{-13} \).