Question

There are 3 consecutive even integers that have a sum of 144. What is the least of the 3 integers?

Answer 1

To solve for the three consecutive even integers that have a sum of 144, we can define the integers in terms of a variable.

Let the first even integer be \( x \). Then the next two consecutive even integers can be expressed as \( x + 2 \) and \( x + 4 \).

Now, we can set up the equation for their sum:

\[
x + (x + 2) + (x + 4) = 144
\]

Combining like terms, we get:

\[
3x + 6 = 144
\]

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

\[
3x = 144 - 6
\]
\[
3x = 138
\]

Now, divide both sides by 3:

\[
x = \frac{138}{3} = 46
\]

So, the first even integer is \( 46 \). The other two consecutive even integers are:

\[
x + 2 = 46 + 2 = 48
\]
\[
x + 4 = 46 + 4 = 50
\]

Thus, the three consecutive even integers are 46, 48, and 50.

The least of these integers is:

\[
\boxed{46}
\]