Question 865115
Let's call those numbers {{{n-2}}} , {{{n}}}, and {{{n+2}}} .
(Those are 3 numbers, with each greater than the one before by 2.
The approach could be the same for a problem involving 3 consecutive even numbers.
For this problem, when we solve for {{{n}}} , {{{n}}} needs to be an odd number.
If we found a value for {{{n}} that is not an odd number,
either we made a mistake in the calculations,
or the problem has no solution).
 
{{{(n-2)+n+(n+2)=3n}}} is the sum of all three consecutive odd integers.
{{{n(n+2)}}} is the product of the larger two consecutive odd integers.
The problem states that
{{{(n-2)+n+(n+2)=n(n+2)-42}}} or {{{3n=n(n+2)-42}}} .
 
{{{3n=n(n+2)-42}}} --> {{{3n=n^2+2n-42}}} --> {{{n^2+2n-42-3n=0}}} --> {{{n^2-n-42=0}}}
That last equation is easy to solve by factoring:
{{{n^2-n-42=0}}} --> {{{(n+6)(n-7)=0}}} --> {{{system(either,n+6=0,or,n-7=0)}}} --> {{{system(either,n=-6,or,n=7)}}} .
Since we are expecting {{{n}}} to be an odd integer, {{{highlight(n=7)}}} ,
and the three consecutive odd integers are
{{{n-2=7-2=highlight(5)}}} , {{{n=highlight(7)}}} , and {{{n+2=7+2=highlight(9)}}} .