# If the sum of three consecutive even integers is 66 what are the integers?

Mar 31, 2018

$20 , 22 , 24$

#### Explanation:

Take the first integer as $x$. Since we are talking about the next even integer, it will come when you add two to $x$.

For example, let's say that you have $2$ as the value of $x$. The next even integer is $4$, which is $x + 2$, and the next is $6$, which is $x + 4$.

So eventually, the equation is

$\left(x\right) + \left(x + 2\right) + \left(x + 4\right) = 66$

$3 x + 6 = 66$

$3 x = 66 - 6$

$3 x = 60$

x = 60 ÷ 3

$x = 20$

So

• $x = 20$
• $x + 2 = 22$
• $x + 4 = 24$
Mar 31, 2018

So the three numbers are: 20; 22; 24

#### Explanation:

You have to 'jump' an odd number to get to the next even number. So even numbers occur every second place (counting in 2's)

Set the first even number as $n$
Then the second is $n + 2$
The third is $n + 4$

$\left(n\right) + \left(n + 2\right) + \left(n + 4\right) = 3 n + 6 = 66$

Subtract 6 from both sides

$3 n = 60$

Divide both sides by 3

$n = 20$

So the three numbers are: 20; 22; 24

Mar 31, 2018

Let us assume the first even number to be color(purple)(x.

So the other to Consecutive even numbers would be:

color(purple)(x+2 and x+4.

So, according to the question,

color(green)(->(x)+(x+2)+(x+4)=66

$\to x + x + x + 2 + 4 = 66$

$\to 3 x + 6 = 66$

$\to 3 x + \cancel{6} - \cancel{6} = 66 - 6$

$\to 3 x = 60$

$\frac{\cancel{3} x}{\cancel{3}} = \frac{60}{3}$

color(magenta)(x=20

And so, the three consecutive even numbers which sum up to 66 :

color(red)(1. x=20

color(red)(2. x+2=22

color(red)(3. x+4=24

As a check:

The sum of all three should be $66$

color(blue)(=20+22+24

color(blue)(=66

Hence it is proved that the 3 no.s are color(Darkorange)(20, 22 and 24

~Hope this helps! :)