Question 812811
the sum of four consecutive even integer is 2 more than five times the first integer.  Find the smallest integer.
<pre>
First (smallest) consecutive even integer = n
Second consecutive even integer = n+2
Third consecutive even integer = n+4
Fourth consecutive even integer = n+6

n+(n+2)+(n+4)+(n+6) = 5n+2

      n+n+2+n+4+n+6 = 5n+2

              4n+12 = 5n+2
         
                 10 = n

Checking:  The 4 consecutive even integers are 10, 12, 14, and 16.

Their sum is 10+12+14+16 = 52.

Five times the first is 5×10 = 50

2 more than 50 is 52.

So the answer is correct.

Edwim</pre>