Question 716439
Let's call the smallest of those numbers {{{n}}} and say that it is an even number.
 
THE WAY TO WRITE THE ANSWER TO THE PROBLEM:
{{{n+(n+2)+(n+4)=78}}}
{{{n+n+2+n+4=78}}}
{{{3n+6=78}}}
{{{3n=78-6}}}
{{{3n=72}}}
{{{n=72/3}}}
{{{highlight(n=24)}}}
 
THE EXPLANATION (in English):
You do not need to write this.
I know I use lots of words, but I am hoping it is an understandable explanation.
We do not need to make a fancy expression to say that it is an even number for this kind of problem.
We can just call it {{{n}}} and trust that it will be an even number.
We will find at least one even number as a solution.
Otherwise the problem has no solution, and it was an unfair trick question.
Since {{{n}}} is even, the even number that follows {{{n}}} is {{{n+2}}}.
(The number in between is {{{n+1}}} which would be an odd number).
The even number that follows {{{n+2}}} is {{{(n+2)+2=n+4}}}.
The sum of those 3 consecutive even numbers is
{{{n+(n+2)+(n+4)}}}
When adding numbers, the commutative and associative properties tell us that we get the same result adding them in any order, grouping them in any way we want.
I did not even need to write those brackets that separate those even numbers.
I can write the sum rearranged in any order, with or without brackets.
{{{n+(n+2)+(n+4)=n+n+2+n+4=(n+n+n)+(2+4)=3n+6}}}