Question 827785
WITH PENCIL AND PAPER (SLOW):
If I was going to use pencil and paper, I would start by assigning a variable name or an expression to one of those integers.
An even integer can be called {{{n}}} or {{{2n}}} .
IF you start with
{{{2n}}}= the greatest of the 5 consecutive even integers, the other even integers are
{{{2n-2}}} , {{{2n-4}}} , {{{2n-6}}} , and {{{2n-8}}} .
The sum of those 5 consecutive even integers is
{{{2n+(2n-2)+(2n-4)+(2n-6)+(2n-8)=120}}}
Solving:
{{{2n+(2n-2)+(2n-4)+(2n-6)+(2n-8)=120}}}
{{{2n+2n-2+2n-4+2n-6+2n-8=120}}}
{{{2n+2n+2n+2n+2n=120+2+4+6+8}}}
{{{10n=140}}}
{{{n=140/10}}}
{{{n=14}}}
{{{2n=2*14}}}
{{{highlight(2n=28)}}}
 
FASTER MENTAL MATH:
(I can think my way to the solution fast, but explaining it takes time).
I know that consecutive integers are evenly spaced.
(Each one is 2 more than the one before; they are an arithmetic sequence).
For an arithmetic sequence, if I have 5 of the terms (or 7 of them, or 9, or any odd number of them), the middle term is the average and the median.
So the sum of 5 consecutive even integers is 5 times the middle integer (the third integer).
So 5 times that integer is 120, meaning that the middle integer is 24, and the next to even integers are 26 and {{{highlight(28)}}} .
The greatest of the 5 even integers is {{{highlight(28)}}} .