A set of 4 consecutive odd integers sums to 40. Which integers are they?

Answer 1

Let the four consecutive odd integers be represented as \( x, x+2, x+4, \) and \( x+6 \).

We know that the sum of these integers equals 40:

\[ x + (x + 2) + (x + 4) + (x + 6) = 40 \]

Simplifying the left side:

\[ 4x + 12 = 40 \]

Now, we will subtract 12 from both sides:

\[ 4x = 40 - 12 \] \[ 4x = 28 \]

Next, divide both sides by 4:

\[ x = 7 \]

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

1. \( x = 7 \)
2. \( x + 2 = 9 \)
3. \( x + 4 = 11 \)
4. \( x + 6 = 13 \)

Thus, the four consecutive odd integers are:

\[ 7, 9, 11, 13 \]

To confirm, we can check the sum:

\[ 7 + 9 + 11 + 13 = 40 \]

Therefore, the integers are \( \boxed{7, 9, 11, 13} \).