Question

There are 4 consecutive even integers that sum to 28. What is the value of the least integer?

Answer 1

Let:

2n = 1st integer

2n + 2 = 2nd integer

2n + 4 = 3rd integer

2n + 6 = 4th integer

So:

$\begin{gathered} 2n+(2n+2)+(2n+4)+(2n+6)=28 \\ 8n+12=28 \\ n=\frac{16}{8} \\ n=2 \end{gathered}$

Therefore the least value is :

$2n=2(2)=4$