Hi, I’m Rob. Welcome to Math Antics.
In math, even when there’s just one right answer to a particular problem,
there’s often many different ways that you could have arrived at that answer.
Take division for example.
Suppose you need to find the answer to the problem 16 divided by 5.
One way we could get the answer is to model the problem with a simple drawing
(or even physical objects).
Like, you could draw 16 circles
and then see what you’d get if you divided them them into 5 equal groups.
Notice that we ended up with 3 circles in each group but we have 1 left over.
That means that we figured out that 16 divided by 5 equals 3 with a remainder of 1.
Of course, there are ways to divide up remainders too using decimals,
but in this video, we’re just going to accept remainders in our answers.
Another method we could use to calculate 16 divided by 5 would be the standard division algorithm.
That’s what you learned if you watched these previous Math Antics videos.
And in this video, whenever I say “standard algorithm” or “standard method”, that’s what I mean.
For that method, we’d set up the problem like this and then go digit-by-digit first asking,
“How many 5s will go into 1?”
Since the answer is 0, we combine the 1 with the next digit and ask,
“How many 5s will go into 16?”
From the multiplication table, we know that 3 x 5 = 15,
so we put a 3 in our answer line above the 6.
Then we multiply 3 x 5 to get 15 and put that below our dividend so we can subtract to get the remainder.
16 - 15 = 1, so we end up with 3 with a remainder of 1 just like we did with our other method.
That’s cool, but there’s another method for doing division that you may or may not have heard about before.
It’s called the Method of Partial Quotients.
Like most methods, the Method of Partial Quotients has advantages and disadvantages.
One of its main advantages is that you don’t need to know a lot of different multiplication facts to use it.
For example, when we used the standard method to divide 16 by 5,
it really helped us out to know that 3 x 5 = 15.
But what if we hadn’t memorized that multiplication fact yet?
What if we only knew that 1 x 5 = 5 and that 2 x 5 = 10?
With the Partial Quotients Method, that would be enough to get us the right answer.
Here’s how it works…
You start out exactly like you do with the standard division algorithm.
But instead of asking how many times 5 will go into 16
and trying to get the answer to be as close as possible with out going over.
We’re just going to pick ANY multiple that we’re sure will go into it. …like 2.
We know that two 5s equals 10 and it’s easy to see that 10 is less than 16,
so we can start by estimating that the answer will be 2
(even though we know that’s not quite right).
But here’s the important difference…
We’re not going to put that 2 up in the answer line.
Instead, we’re going to write it off to the right side of our work like this
(you’ll see why in a minute).
Then we’ll continue like we would with the standard algorithm:
2 x 5 = 10 so we’ll put the 10 below the 16 and subtract it to see what the remainder is.
16 minus 10 equals 6. Now what?
Well… since the remainder is greater than the divisor, we know that we can keep dividing.
In other words, the initial answer we guessed (2) was too small and needs to be adjusted with another step.
So now we ask, how many 5s will go into the remainder and it’s pretty easy to see that the answer will be 1.
Again, instead of writing that 1 in the answer line,
we’ll put it here on the right side of our work below our first estimate.
Then we continue as before.
1 x 5 = 5 which we’ll put below the 6 to subtract it, and 6 - 5 gives us a new remainder of 1.
Now the remainder is less than the divisor, so we’re almost finished.
We just need to finalize our answer by combining the answers we got along the way.
And we need to combine them because they were just ‘partial answers’, or ‘partial quotients’.
Get it? “Quotient” is just the fancy name for the answer to a division problem.
That’s why we didn’t write the answers up in our answer line.
We needed to wait until we got a remainder that was less than the divisor
so we were sure we had gotten all the partial answers that we needed.
Now if we add up those partial answers, 2 + 1 = 3,
we’ll get the final answer that we can write either down below like this
or up at the top in the answer line if you prefer.
But don’t forget the remainder of 1.
Just like with the standard division method, that’s part of the answer too.
As you can see, the Partial Quotients Method gave us the exact same answer
as the other two methods of division we tried.
You might be thinkig, that method seemed like a bit more work than the standard method, didn’t it?
Well yes… but that’s because in the standard method,
we were helped out by the fact that we already knew all our multiplication facts for multiples of 5.
That meant we could easily determine that 3 was the best choice for the answer.
We knew that three 5s would be very close to 16 but that four 5s would be too many.
But what if we were given a much tougher division problem like this one?
6428 divided by 35.
I don’t know about you, but I don’t have my multiplication table memorized all the way out to the 35s.
So what will we do?
Should we calculate all the multiples of 35 we that we might need ahead of time?
Well, with the Partial Quotients Method, we won’t need to
because that method lets us just pick an answer that we think be close enough for each step,
and then we can just do more steps as needed to adjust that answer as we go along.
For example, I have no idea exactly how many times 35 will go into 6428,
but I do know that it will go in at least 100 because 100 x 35 = 3500.
And since that multiple was SO easy to calculate,
I’m going to choose that as my first partial answer.
I’ll write 100 off to the side and then since 100 x 35 = 3500,
I’ll subtract that from 6428 to get a remainder of 2928.
That’s a HUGE remainder when compared to our divisor, isn’t it?
But that’s totally okay in the Partial Quotients Method
because it just means that we have more partial answers to figure out.
So what should our next estimate (or guess) be for a partial answer?
Well, we can’t do 100 again since 3500 is greater than the remainder,
so we need something smaller.
10 x 35 = 350. That’s definitely smaller,
but if we use values that are too small,
it will take us a lot longer to chisel the remainder down to an acceptable level.
So how about if we double that?
If 10 x 35 = 350 then 20 x 35 would be 700.
700 is certainly closer to the size of our remainder,
but it looks like if we double it again we can get even closer,
and doubling 700 is easy… it’s just 1400.
So if twenty 35s gives us 700, then forty 35s will give us 1400.
1400 is less than our remainder, but it’s close,
so let’s use 40 as our next partial answer by writing to on the right hand side like this.
40 x 35 = 1400 and 2928 - 1400 = 1528.
Cool! Our remainder has gotten a lot smaller,
but it’s still much bigger than the divisor, so we need to keep going.
The good news is that it looks like the last partial quotient we used
will also work for this step, so let’s use it again.
That means we’ll write another 40 off to the side.
40 x 35 = 1400 and 1528 - 1400 = 128.
Well, we’re definitely getting closer now
since our remainder is just a little bigger than our divisor.
I’m not sure exactly how many times 35 will go into 128,
but I know it will go in at least twice since 2 x 35 = 70,
so that will be my next guess for a partial answer.
2 x 35 = 70 and 128 - 70 = 58.
D’oh… that remainder is still bigger than the divisor,
so it looks like we could divide one more 35 into the reminder.
So let’s pick 1 as our next (and hopefully final) partial answer.
1 x 35 = 35 and 58 - 35 = 23
Great! Our remainder is finally less than our divisor, so we’re done dividing.
But, what’s our final answer?
To get it, we need to add up all the partial answers:
100 + 40 + 40 = 180 and 180 + 2 + 1 = 183.
So our final answer is 183 r 23
Wow! That still seamed like a lot of work
and you may be wondering if the Partial Quotients Method was really a better idea
than just using the traditional division algorithm.
What was the advantage?
Well, one advantage is that we were able to solve the problem
with a set of multiplication facts that was pretty easy to calculate.
They were easy because we got to choose convenient multiples.
It’s pretty easy to multiply things by 10 or by 2.
If we had used the traditional method, we would have had to know the multiplication facts
3 x 35 = 105 and 8 x 35 = 280.
Sure, those wouldn’t have been that hard to calculate,
but it was very hard to tell at the beginning of the problem
that those were the exact set of multiplication facts that we needed to know.
We probably would have had to do a lot of trial and error to figure that out.
So even though the Partial Quotients Method usually involves more steps,
it also offers you a lot of flexibility.
With the standard division method,
there’s basically just one path that you follow to get to the answer,
because for each step, you’re always supposed to find the exact multiple
that results in the smallest possible remainder.
But with the Partial Quotients Method,
the size of the remainder doesn’t matter until the very end,
so you’re free to choose the easiest multiples you can find for each step.
In other words, you can choose your own path to get to the answer.
But choose wisely, so you don’t make the problem a lot longer than necessary.
To see what I mean, let’s give two totally different students the same division problem
and ask them both to use the Partial Quotients Method to do it.
The problem is 1276 divided by 26.
Let’s see which path they choose to get to the answer.
Ready? Go!
Interesting… They both got the same answer,
but one student solved the problem in 7 steps,
while the other solved it in only 4 steps.
What made the difference?
Well, from their scratch work,
it looks like each student figured out that 2 x 26 = 52 and therefore 20 x 26 = 520.
The first student decided that knowing those two multiples was good enough
and began chipping away at the dividend with them.
Their first partial answer was 20 so they subtracted 520 from the dividend to get a remainder of 756.
They realized they could do that again for the next step which brought the remainder down to 236
The next partial answer they chose was 2, which let them subtract 52,
and they did that a total of 4 times until the remainder
got small enough to subtract their last multiple of 26.
Last of all they added up the partial answers they got
to get a final answer of 49 with a remainder of 2.
But it looks like the second student realized that
if he calculated a couple bigger multiples at the being
(specifically 4 x 26 = 104 and 40 x 26 = 1040)
he could use them to chip away at the dividend even faster.
In the very first step he was able to subtract 1040 from the dividend,
so the remainder got smaller more quickly.
And knowing that 4 x 26 = 104 let him get the remainder down to 28 in just two additional steps.
So a little extra multiplication work at the beginning
ended up saving him a lot of subtraction along the way.
That also meant he had fewer partial answers to add up to get the final answer.
But even though they took different paths,
both students did a great job using the Partial Quotients Method
since they each got the correct answer!
As you can see,
the key to getting good results with the Partial Quotients Method is to choose a set of multiples
that will help you take big but easy ‘bites’ out of the dividend with each step.
But it can be hard to decide which ones you should calculate at the beginning.
Since multiplying the divisor by 1 or 10 is super easy, those are an obvious start.
And it’s usually not too hard to multiply by 2 and 5 to get those multiples.
But if you’re having trouble deciding which multiples to use, don’t worry too much about it.
Just pick 2 or 3 and see what happens.
Good evening monsieur, have you decided what multiples you’d like this fine evening?
Ya know what… Surprise me!
Ah… Excellent choice monsieur.
The good news is the more you use the Partial Quotients Method,
the better you’ll get at choosing multiples that make the process as easy as possible.
But it’s important to remember that some division problems are just inherently harder than others
and some may be better suited for the Partial Quotients Method than others.
The important thing is that you understand the process.
Alright, that’s it for this video.
Now you have one more tool in your math toolbox for solving division problems.
If you’ve memorized your multiplication table,
you may just want to use the traditional division method for problems that have a divisor of 12 or less.
But for bigger divisors, the Partial Quotients Method can really come in handy…
that is if you don’t have a calculator of course.
Remember, the best way to get good at math is to practice,
so try solving several problems on your own.
As always, thanks for watching Math Antics and I’ll see ya next time.
And did monsieur save a remainder for dessert?
No… no thank you… I’m trying to watch my figures.
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