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

1 answer

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 \( \boxed{-13} \).