# There are three consecutive even integers. If twice the first integer added to the third is 268,258, what are all three integer?

Sep 19, 2017

89418 , 89420 , 89422

#### Explanation:

Let $n$ be any positive integer. Then $2 n$ is a even integer (multiplying by 2 makes the number even, because 2 will be a factor of that number and all even numbers are divisible by 2).

$2 n + 2$ is the next integer and $2 n + 4$ the next.

(Adding 2 to an even number gives you the next even number, adding 2 again the next even number and so on)

So we have three consecutive integer:

$2 n , \left(2 n + 2\right) , \left(2 n + 4\right)$

Twice the first added to the third is $268 , 258$

$2 \left(2 n\right) + \left(2 n + 4\right) = 268258$

Solving for $n$:

$4 n + 2 n + 4 = 268258$

$6 n = 268254$

$n = \textcolor{b l u e}{44709}$

So we have:

$2 \left(\textcolor{b l u e}{44709}\right) , \left(2 \left(\textcolor{b l u e}{44709}\right) + 2\right) , \left(2 \left(\textcolor{b l u e}{44709}\right) + 4\right) \to$