Partial quotients

When you reach for it

Lalit makes friendship bracelets. Each bracelet takes 6 glass beads, and he has 1{,}674 beads. How many bracelets can he make?

This is a “how many groups” question: how many groups of 6 fit in 1{,}674? That is the division 1{,}674 \div 6 (see What division means, where division asks either how many in each group or how many groups). You could count out 6 beads, then 6 more, then 6 more. But that takes hundreds of steps, and no times-table fact gives the answer.

So take away many groups at once. Ten groups of 6 is 60 beads. A hundred groups of 6 is 600 beads. Those are easy to find, because each one is a basic fact and a shift of places (see Multiplying by tens and hundreds). Take away a big easy chunk, then another, until no group of 6 is left. Then add up how many groups you took. Each count is called a partial quotient, since it is one part of the whole answer.

The words: in 1{,}674 \div 6 = 279, the number being divided, 1{,}674, is the dividend. The number it is divided by, 6, is the divisor. The answer, 279, is the quotient. If some is left over that is too small to make one more group, that amount is the remainder.

Reach for partial quotients whenever the dividend is too big for a basic fact.

The procedure

This lesson takes four days, one part a day.

Part 1: taking away easy chunks

Start with all 1{,}674 beads. A hundred bracelets use 100 \times 6 = 600 beads, and 600 is less than 1{,}674, so Lalit can surely make 100 bracelets. Take those beads away:

There are still more than 600 beads, so take away another 100 bracelets’ worth:

Now 600 is too many. Try tens of bracelets. Since 7 \times 6 = 42, seventy bracelets use 70 \times 6 = 420 beads: 7 groups of 6 tens is 42 tens. And 420 is less than 474:

Last, 54 beads is a basic fact: 9 \times 6 = 54, so 9 more bracelets, with nothing left.

Lalit made 100, then 100, then 70, then 9 bracelets. Add the groups:

So 1{,}674 \div 6 = 279. He can make 279 bracelets.

People record this work in a ladder. The left side shows what is left after each step. The right side, past a line, counts the groups taken.

Why does adding the groups give the answer? Every bead was used in one of the chunks, so the chunks add back to the whole:

The step from the second line to the fourth is the The distributive property: 6 times each part, added up, is the same as 6 times the sum of the parts. Splitting the dividend into chunks that are each 6 times something is that property read backward. And since 6 \times 279 = 1{,}674, the quotient is 279.

The same chunks make a picture. Think of 1{,}674 as the area of a rectangle that is 6 tall. Its width is unknown, and the width is the quotient. Each chunk taken away is a piece of the rectangle: 600 is a piece 6 tall and 100 wide. Lay the pieces side by side and read the width across the top.

This is the area model you met in Partial products, run the other way. There you knew both sides and found the area. Here you know the area and one side, and you build the other side piece by piece.

Part 2: bigger chunks or smaller chunks

Lalit took 100 groups twice. He could have taken 200 groups at once, since 200 \times 6 = 1{,}200 (two groups of 6 hundreds is 12 hundreds), and 1{,}200 is less than 1{,}674:

Three steps instead of four, and the same quotient. Someone else might use only tens for a while, taking 60 again and again. That is many more steps, but it still ends at 279.

Any sensible chunks give the same quotient. Here is why. Whatever chunks you take, each one is 6 times its count of groups, and together the chunks use up all 1{,}674 beads. So 6 \times (\text{sum of the counts}) = 1{,}674, by the distributive property again. Only one whole number times 6 makes 1{,}674, so the sum of the counts is always 279.

“Sensible” means two things. A chunk must be a whole number of groups of the divisor, so you can count it. And it must be no bigger than what is left, since you cannot take away beads you do not have.

So which chunks are best? The biggest easy chunk at each step gives the fewest steps. Easy chunks come from a basic fact and a shift: 1, 2, ..., 9 groups, 10, 20, ..., 90 groups, and 100, 200, ..., 900 groups. Five groups and fifty groups are friendly too, being half of ten and half of a hundred. A good habit is to ask “how many hundreds of groups fit?”, then “how many tens of groups?”, then “how many groups?”. That picks one chunk per place value.

If a chunk turns out too big, nothing breaks. You find out when you try to subtract it, and you pick a smaller one.

Part 3: remainders

Tui is setting out flower pots at a garden store. Each shelf holds 8 pots, and there are 947 pots. How many shelves can be filled?

That is 947 \div 8. A hundred shelves hold 800 pots, so take those away: 947 - 800 = 147. Ten shelves hold 80: 147 - 80 = 67. Since 8 \times 8 = 64, eight more shelves hold 64: 67 - 64 = 3.

Now 3 pots are left, and 3 is less than 8. No more full shelves can be made, so the ladder stops. The groups add to 100 + 10 + 8 = 118. So 118 shelves are filled and 3 pots are left over. Write

read “947 divided by 8 is 118, remainder 3”.

The ladder stops at the first number less than the divisor, and that number is the remainder. It can be any whole number from 0 up to one less than the divisor. A remainder of 0 means the divisor fits exactly, as it did for the beads. A remainder as big as the divisor would mean one more group still fits, so the work is not done.

What the 3 means depends on the story: 3 pots on a shelf of their own, 3 pots in a box, or a reason to round up. That is a question about the context, taken up in Remainders and what they mean. The arithmetic gives 118 and 3 either way.

Four-digit dividends work the same way, with chunks of up to nine hundred groups (Example 6). A quotient can have a zero in it too. When the hundreds of groups leave less than ten groups, you take no tens at all, and the tens digit of the quotient is 0 (Example 5).

Part 4: the area model, and checking by multiplying back

Liv divides 5{,}203 by 7. The rectangle is 7 tall, and she builds its width from the left.

• Hundreds of groups: 7 \times 700 = 4{,}900 fits in 5{,}203, and 7 \times 800 = 5{,}600 does not. Take away 4{,}900, leaving 303. The first piece is 700 wide.
• Tens of groups: 7 \times 40 = 280 fits in 303, and 7 \times 50 = 350 does not. Take away 280, leaving 23. The next piece is 40 wide.
• Groups: 7 \times 3 = 21 fits in 23. Take away 21, leaving 2. The last piece is 3 wide.

The width is 700 + 40 + 3 = 743, and 2 is left over. So 5{,}203 \div 7 = 743 \text{ R } 2.

The picture also says how to check. The pieces are the partial products of 743 \times 7 (see Partial products), so multiply back and add the remainder:

That is the dividend, so the division is right. In general,

and the remainder must be less than the divisor. Both parts matter. The answer 742 \text{ R } 9 also passes the first test, since 742 \times 7 + 9 = 5{,}203, but 9 is not less than 7: one more group of 7 was still sitting in the leftovers.

A quick estimate catches a lost place before you even multiply. 5{,}203 is close to 4{,}900, and 4{,}900 \div 7 = 700, so the quotient should be a little more than 700. It is.

Worked examples

Boundaries

• The divisor cannot be zero. A group of 0 beads uses up nothing, so you could take away groups forever and never finish. Division by 0 has no answer.
• A chunk must be a whole number of groups of the divisor. Each amount taken away is the divisor times its count. If it is not, the count on the right is not the number of groups taken.
• A chunk cannot be bigger than what is left. These are whole numbers: you cannot take away more than you have. A chunk that is too big just means trying a smaller one.
• The ladder stops when what is left is less than the divisor. That amount is the remainder, so a remainder is always less than the divisor. If what is left is still as big as the divisor, another group fits and the work is not finished.
• Which chunks you take changes the steps, not the answer. For whole numbers, the quotient and remainder are the same for every correct ladder.
• This lesson divides by one digit. The same ladder works for any divisor, but finding the chunks for a larger divisor needs a trial multiple at each step (see Dividing by a two-digit number).

Where this goes next

Partial quotients show where every part of a quotient comes from. When you choose the biggest chunk for each place value, one chunk of hundreds of groups, one of tens and one of ones, the right column holds exactly one digit of the answer per place. Writing only those digits, in place, gives the short layout of Dividing by a single digit, and then the standard long division algorithm, which does the same work one place at a time.

A remainder in a story needs a decision: drop it, round up, or share it out. That is Remainders and what they mean.

Exercises

Part 1: taking away easy chunks

1. Find 168 \div 7 using chunks of ten groups.
2. Find 865 \div 5 with partial quotients. Record your work in a ladder.
3. Find 1{,}248 \div 3 with partial quotients.
4. Lalit has 912 beads, and each bracelet takes 8. How many bracelets can he make?
5. Tui divides 2{,}142 by 7 and takes away 2{,}100 first. How many groups of 7 is that? What is left? Finish the division.
6. Liv divides 1{,}092 by 4 and takes away 800, 200, 80 and 12. How many groups did she take each time, and what is the quotient? Write 1{,}092 as 4 times a sum.

Part 2: bigger chunks or smaller chunks

7. Find 738 \div 6 two ways: once using only chunks of 100 and 10 groups (and ones at the end), and once using the biggest chunk you can at each step.
8. Find 2{,}575 \div 5, starting with a chunk of 500 groups.
9. Liv wants to start 1{,}516 \div 4 by taking away 400 groups. Can she? Find the quotient.
10. Tui finds 3{,}330 \div 9 by taking away 900 three times and then 630. What is the quotient? What single chunk could replace her first three?
11. A garden shelf holds 7 flower pots. How many shelves do 1{,}449 pots fill?
12. For 896 \div 8, Lalit’s groups are 100, 10 and 2. Liv’s groups are 50, 50, 10, 1 and 1. Find each quotient. Why do they agree?

Part 3: remainders

13. Find 359 \div 3. Give the quotient and the remainder.
14. Find 746 \div 7.
15. Find 2{,}893 \div 6.
16. Find 5{,}003 \div 8.
17. Liv has 1{,}237 flower pots and puts 9 on each garden shelf. How many full shelves are there, and how many pots are left over?
18. Tui says 214 \div 5 = 41 \text{ R } 9. What is wrong with her answer? Fix it.

Part 4: the area model and checking

19. Draw an area model for 1{,}365 \div 3 with pieces 400, 50 and 5 wide. What are the areas of the pieces, and what is the quotient?
20. An area model is 8 tall. Its pieces have areas 3{,}200, 560 and 48. What division does it show, and what is the quotient?
21. Is 2{,}461 \div 7 = 351 \text{ R } 4 right? Check by multiplying back.
22. Lalit says 4{,}250 \div 6 = 78 \text{ R } 2. Check by multiplying back, then find the right answer.

Reasoning

23. Explain why two people who take different chunks for the same division must get the same quotient.
24. Explain why the remainder in a division by 6 can never be 6 or more.
25. Lalit says 2{,}400 \div 6 needs no ladder at all. Use a basic fact and place value to explain why, and give the quotient.