Whats three consecutive even integers whose sum is 54?

2 Answers
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, color(white)("d")n+2, color(white)("d")n+4

Adding these up (their sum) we have:

(n) +(n+2)+(n+4)larrThe brackets just demonstrate the
color(white)("dddddddddddddddddddddd")grouping. They serve no othercolor(white)("dddddddddddddddddddddd") purpose.

Their sum is 3n+6=54

Subtract 6 from both sides
color(white)("dddddddddddd.")3n=48

Divide both sides by 3
color(white)("ddddddddddddd.")n=16
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
First number color(white)("d..")->16
Second number ->18
Third number color(white)(".d")->ul(20larr" Add as a check")
color(white)("ddddddddddddddd")54

Mar 13, 2018

16, 18, and 20

Explanation:

Let x = our first even integer.

We can write the problem as:
x + (x+2) + (x+4) = 54

Add like terms:
3x + 6 = 54

Rearrange and solve for x:
3x = 54 - 6
3x = 48
x = 16

Therefore our three consecutive even integers are 16, 18, and 20.