Question 1053483
Let {{{n}}} , {{{n+2}}} , {{{n+4}}} and {{{n+6}}} be the four consecutive even integers.
So,
{{{n+(n+2)}}}= the sum of the first and second integers,
{{{2(n+(n+2))}}}= two times the sum of the first and second integers,
{{{3(n+6)}}}= three times the fourth integer, and
{{{3(n+6)+40}}}= 40 more than three times the fourth integer.
The problem says that
{{{2(n+(n+2))=3(n+6)+40}}} ,
and that is the equation we have to solve.


{{{2(n+(n+2))=3(n+6)+40}}}
{{{2(2n+2)=3n+18+40}}}
{{{4n+4=3n+58}}}
{{{4n-3n=58-4}}}
{{{highlight(n=54)}}}
So the four consecutive even integers are
{{{highlight(54)}}} , {{{highlight(56)}}} , {{{highlight(58)}}} and {{{highlight(60)}}} .
 
Verification:
The sum of the first and second integers is {{{54+56=110}}} .
Two times the sum of the first and second integers is {{{2*110=220}}} .
Three times the fourth integer is {{{3*60=180}}} .
Forty (40) more than three times the fourth integer is {{{180+40=220}}} .