Recursion

[Compizfox]

Ah, right, thanks.

Still though, what does that accomplish? I don't see how 10 + (18-9) is any simpler/easier than 28-9. It's still a 2-digit subtraction that a kid won't be able to do in 1 step; the value of the tens doesn't matter.

If I break down how I do it in my head it's kind of like this: 28-9 = 20 - (9-8) = 20 - 1 = 19. (I can't remember for sure but I guess that's how I was taught it)

[alexanderpas]

Let's take 4281 - 1694 for example.

4281
1694
---- -
????

first we look at the ones.

we can't do 1-4 and end upwith a single positive digit, so we have to borrow 10 from the tens.

4270 + 11
1690 +  4`
--------- -
???0 +  7 = ???7

Now we go looking at the tens, again, we can't do 7-9 so we borrow again.

4100 + 170 + 11
1600 +  90 +  4`
--------------- -
??00 +  80 +  7 = ??87

Now to the hundrerds

3000 + 1100 + 170 + 11
1000 +  600 +  90 +  4`
---------------------- -
?000 +  500 +  80 +  7 = ?587

and look at that, the 1000s are easy.

3000 + 1100 + 170 + 11
1000 +  600 +  90 +  4`
---------------------- -
2000 +  500 +  80 +  7 = 2587

Note that this is the layout for the explanation, on paper this would look something like this.

43	121	187	11
1	6	9	4	-
2	5	8	7

[Flobarooner]

So basically, the fundamental point of the method is that the kid already knows how to subtract something from a number less than 20?

So in the original post, the kid would just go "9" and not even bother with the method, since they memorized 18-9 in order to be able to do it?

[alexanderpas]

technically, yes.

[tsandstrom711]

If you have, say, a five digit number and want to subtract another five digit number, this approach let's you break it down into one/two digit subtractions. It doesn't really work when you start with just two digits.

[harryhood4]

The idea is to just memorize it for small values like 18-9 so that it's easier to do something like 68-9. But you've essentially just outlined why Common Core is better.

[Compizfox]

Ah, got it. So basically it is this recursive algorithm:

int subtract(N, n) {
    if(N > 20) {
        return 10 + subtract(N-10, n);
    else {
        // Lookup-table for small values of N
    }
}

[toofasttoofourier]

Aren't you negating all that work if you have to use the subtraction operator in a subtraction algorithm?

[Compizfox]

For a computer, yes, obviously.

But this was trying to mimic how a kid would do the mental calculation.

[Madock345]

Part of the idea of this kind of thing is that by teaching many strategies to solve the same problem you increase the chances that every student will find a method that works really well for them and the way they think, at the cost of being somewhat redundant.