Yeah, I would be unable to solve it, given that I'm pretty sure there is no solution if we assume that "consecutive" implies integers (and I'd be curious what 'consecutive' means if we don't).
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).
You're right, my mistake. That "five higher" should have been "four higher".
The thing is that your solution is the algebraic solution. You're simplifying an equation by balancing both sides, when you subtract four. Just from our perspective, you simplify to a still difficult state, instead of the easiest possible state.
Algebra is just a way of formally stating what you did, and then offering some simplifications that speed up the process. Or offering more powerful methods that making solving more difficult problems easier.
'The thing is that your solution is the algebraic solution'
I agree. That's why I said
'Really, the trick mention(ed) ... is just a rephrasing of the algebra'
Just because you do number juggling in your head rather than write it formulaically doesn't make it an inferior technique, or make it not algebra, which was my point.
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).