We can express this question as follows:
[tex]n+(n+2)+(n+4)=6[/tex]Now, we can sum the like terms (n's) and the integers in the previous expression. Then, we have:
[tex](n+n+n)+(2+4)=6=3n+6\Rightarrow3n+6=6[/tex]Then, to solve the equation for n, we need to subtract 6 to both sides of the equation, and then divide by 3 to both sides too:
[tex]3n+6-6=6-6\Rightarrow3n=0\Rightarrow n=\frac{3}{3}n=\frac{0}{3}\Rightarrow n=0_{}[/tex]Then, we have that the three consecutive even integers are:
[tex]0+2+4=6[/tex]Therefore, the least integer is 0.
