If instead we made it "2 consecutive numbers and a third number that was five higher than the second number add up to 70" (i.e., 21 + 22 + 27), that's easily solved using 'tricks'. Really, the trick mention (which is what I did in my head, too) is just a rephrasing of the algebra. That is, I would subtract 4 off of the thing that is 5 higher (so that it's now 1 higher; the problem is now the same, get 3 consecutive integers), subtract 4 off of the number I'm trying to get (so 66), so the problem is now 3 consecutive integers that equal 66, and solve the same way (66/3 = 22, so -> 21, 22, 23), and then just add the four back in to the highest (21, 22, 27).