Whats three consecutive even integers whose sum is 54?

Mar 13, 2018

16, 18, 20

Explanation:

To get to the next even number you have to 'jump over' an odd number. So every second number is even if you start from one.

Let the first even number be $n$ so we have:

$n , \textcolor{w h i t e}{\text{d")n+2, color(white)("d}} n + 4$

Adding these up (their sum) we have:

$\left(n\right) + \left(n + 2\right) + \left(n + 4\right) \leftarrow$The brackets just demonstrate the
$\textcolor{w h i t e}{\text{dddddddddddddddddddddd}}$grouping. They serve no other$\textcolor{w h i t e}{\text{dddddddddddddddddddddd}}$ purpose.

Their sum is $3 n + 6 = 54$

Subtract 6 from both sides
$\textcolor{w h i t e}{\text{dddddddddddd.}} 3 n = 48$

Divide both sides by 3
$\textcolor{w h i t e}{\text{ddddddddddddd.}} n = 16$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
First number $\textcolor{w h i t e}{\text{d..}} \to 16$
Second number $\to 18$
Third number $\textcolor{w h i t e}{\text{.d")->ul(20larr" Add as a check}}$
$\textcolor{w h i t e}{\text{ddddddddddddddd}} 54$

Mar 13, 2018

$16$, $18$, and $20$

Explanation:

Let $x =$ our first even integer.

We can write the problem as:
$x + \left(x + 2\right) + \left(x + 4\right) = 54$

$3 x + 6 = 54$
Rearrange and solve for $x$:
$3 x = 54 - 6$
$3 x = 48$
$x = 16$
Therefore our three consecutive even integers are $16$, $18$, and $20$.