Dividing by a single digit

When you reach for it

Meghna runs a card game club. She has 972 game cards to deal out evenly to 4 tables. How many cards does each table get?

That is 972 \div 4, a “how many in each group” question (see What division means). It is too big for a times-table fact. In Partial quotients you took away easy chunks and added them up at the end. This lesson does the same job in a tidier order. It shares the biggest pieces first, then the next biggest, one place at a time, and it writes the answer one digit at a time.

The words, as before: in 972 \div 4 = 243, the number being shared, 972, is the dividend. The number of groups, 4, is the divisor. The answer, 243, is the quotient. What is left over, too small to give every group one more, is the remainder.

Reach for this whenever a whole number of up to four digits is divided by a number from 1 to 9. Each step is a division fact, since the amount being shared at any one place is always small.

The procedure

This lesson takes four days, one part a day.

Part 1: sharing one place at a time, from the left

Think of 972 as base-ten blocks: 9 hundreds squares, 7 ten rods and 2 ones. Meghna shares them among 4 tables, biggest blocks first.

Hundreds. Share 9 hundreds among 4 tables. Each table gets 2 hundreds, since 2 \times 4 = 8, and 3 hundreds each would need 12. That uses 8 hundreds and leaves 1.

Trade. One hundred square cannot be split four ways and stay a hundred. So trade it for 10 ten rods. With the 7 tens already there, that makes 10 + 7 = 17 tens.

Tens. Share 17 tens among 4. Each table gets 4 tens, since 4 \times 4 = 16, and 5 tens each would need 20. That leaves 1 ten. Trade it for 10 ones. With the 2 ones already there, that makes 12 ones.

Ones. Share 12 ones among 4. Each table gets 3, since 3 \times 4 = 12, and nothing is left.

Each table got 2 hundreds, 4 tens and 3 ones, which is 243 cards. So 972 \div 4 = 243.

Here is the same sharing as a table, one row per place. “Share” is how many of that place there are to share, after any trade. “Each” is what each table gets, “Used” is how many were handed out, and “Left” is what is left over, to be traded into the next place.

Three things make this work.

Why start on the left? Big pieces can be traded for small ones, but never the other way while you share. If you shared the ones first, a leftover hundred would later turn into tens and ones, and you would have to share the ones all over again. Starting with the biggest place means anything left can still be passed down to the next place.

Why trade for ten? One hundred is ten tens, and one ten is ten ones. That is what place value means. A trade never changes how many cards there are. It only changes the size of the pieces.

Why is each step a fact? What is left at a place is always less than 4. If 4 or more were left, every table could have had one more. So at most 3 are left, which trade for at most 30 of the next unit. Add the digit already in that place, at most 9, and there are at most 39 to share. Fewer than 40 shared among 4 is a basic fact, and it gives each table at most 9. So each place of the quotient gets one digit, from 0 to 9.

Why is the answer right? Each table got 200 + 40 + 3 cards. Four tables got 4 \times 200, 4 \times 40 and 4 \times 3. By the distributive property,

which is every card. Each product is a fact and a shift of places, as in Multiplying by tens and hundreds. This is partial quotients with the biggest chunk at each place: 200, then 40, then 3.

Part 2: the short record

Writing sentences for every place is slow. The short record keeps only what matters, on a place value chart.

Write the divisor at the left and the dividend under a line, one digit to a column. Over each place, write how many of that place each group gets. When something is left at a place, write it small in front of the next digit. That small digit is the trade: it says “this many were left in the place before, and they are now worth ten times as many here”.

Read the record for 972 \div 4 from the left.

• Over the 9: 9 hundreds shared by 4 is 2 each, with 1 left. The small 1 goes in front of the 7.
• The small 1 and the 7 read together as 17 tens. Over the 7: 17 \div 4 is 4, with 1 left. The small 1 goes in front of the 2.
• The small 1 and the 2 read as 12 ones. Over the 2: 12 \div 4 = 3, with 0 left.

The quotient reads straight off the top: 243. Every digit sits over its place, so the 2 means 2 hundreds without any extra thought.

The same record works for four-digit numbers, starting at the thousands. For any record, the small digit is always less than the divisor, for the reason in Part 1. So the number you read at each place has at most two digits, and the step is a fact you know.

Part 3: a first digit too small to share, and zeros

Ariki stocks a shoe store. He has 5{,}632 pairs of shoes and 8 shoe racks, and every rack should hold the same number of pairs.

Thousands. Share 5 thousands among 8 racks. That is fewer than one each. No rack can get a thousand, so write nothing over the 5 and trade all 5 thousands for 50 hundreds. With the 6 hundreds, that makes 56 hundreds.

Hundreds. Share 56 hundreds among 8: 7 each, since 7 \times 8 = 56, with 0 left.

Tens. Share 3 tens among 8. Again fewer than one each. This time the quotient has already started, so write a 0 over the 3: each rack gets 0 tens. Trade the 3 tens for 30 ones. With the 2 ones, that makes 32 ones.

Ones. Share 32 ones among 8: 4 each, since 4 \times 8 = 32, with 0 left.

So 5{,}632 \div 8 = 704. Each rack holds 704 pairs.

Two things happened that did not happen before.

• A first digit too small. Nothing was written over the 5, since no rack got any thousands. The quotient starts in the hundreds place, so it has three digits, not four. You could write a 0 there, but a number never begins with 0: 0704 is just 704.
• A zero inside the quotient. Over the 3 the digit is 0, and it must be written. It holds the tens place. Without it the 7 and the 4 slide together into 74, and 74 \times 8 is only 592, nowhere near 5{,}632.

The rule that ties these together: once the first digit of the quotient is written, every place to its right gets a digit, even if that digit is 0. The quotient has a digit over every place from the first one shared to the ones.

A quick estimate also catches a lost zero. 5{,}632 is close to 5{,}600, and 56 hundreds shared by 8 is 7 hundreds, so the answer should be about 700. It is.

Part 4: remainders, and checking by multiplying back

Magnhild is in charge of the club’s spare cards. She has 4{,}251 of them to sort into 7 equal piles.

Thousands. 4 thousands among 7 is fewer than one each. Trade them: 42 hundreds. Hundreds. 42 \div 7 = 6, with 0 left. Tens. 5 tens among 7 is fewer than one each, so the tens digit is 0. Trade the 5 tens: 51 ones. Ones. 51 ones among 7 is 7 each, since 7 \times 7 = 49, and 8 \times 7 = 56 is too many. That leaves 51 - 49 = 2.

The ones are the last place. The 2 left over cannot be traded for anything smaller, since there is no place for whole cards below the ones. And 2 is less than 7, so the piles cannot each get one more. The 2 is the remainder:

read “4{,}251 divided by 7 is 607, remainder 2”. Each pile has 607 cards, and 2 cards are left over. What to do with those 2 in a story is its own question, taken up in Remainders and what they mean.

Checking by multiplying back. Multiplication undoes division (see Multiplication and division undo each other). If 7 piles of 607 cards plus the 2 left over make all 4{,}251 cards, the division is right. By the distributive property,

That is the dividend, so the answer checks. In general:

and the remainder must be less than the divisor. Both parts of the check matter. The answer 606 \text{ R } 9 passes the first test, since 606 \times 7 + 9 = 4{,}242 + 9 = 4{,}251. But 9 is more than 7, so every pile could have taken one more card. With both tests passed, the quotient and remainder are the only right pair. That fact, stated for all whole numbers, is the division algorithm.

Worked examples

Examples 1 and 2 go with Part 1, 3 and 4 with Part 2, 5 and 6 with Part 3, and 7 and 8 with Part 4.

Boundaries

• The divisor cannot be zero. Sharing among 0 groups has no meaning, and no number times 0 gives back a dividend other than 0. Division by 0 has no answer.
• This lesson divides whole numbers by one digit, 1 to 9. Dividing by 1 gives the dividend back. Each step is a fact because what is left at a place is less than the divisor, so the number shared after a trade is less than ten times the divisor. A two-digit divisor breaks that: the amount at a place can be much bigger, and each step needs an estimate (see Dividing by a two-digit number).
• Each quotient digit is 0 to 9. A digit of 10 or more would mean the place before had enough left for another full share, so it was too small.
• What is left at a place must be less than the divisor. If it is as big as the divisor, every group can have one more, and the digit over that place goes up by one.
• Once the quotient starts, every place gets a digit. A 0 inside or at the end of the quotient holds a place and must be written. Only zeros before the first digit are dropped.
• The remainder lives in the ones. With whole numbers there is no smaller place to trade into, so what is left at the ones is the remainder, always less than the divisor. Sharing it out as a fraction or as cents is a different question (see Remainders and what they mean).

Where this goes next

The short record is the bones of long division. In the standard long division algorithm the same sharing is written with each product and subtraction shown under the dividend, a layout that keeps working when the divisor has two digits and the trades get too big to do in your head. A remainder in a story needs a decision, and that is Remainders and what they mean.

Exercises

Part 1: sharing one place at a time

1. Find 639 \div 3. Say what each group gets in each place.
2. Find 752 \div 4. Show each trade.
3. Find 915 \div 5 with a sharing table.
4. Find 4{,}862 \div 2.
5. Find 9{,}684 \div 4.
6. Meghna’s card club has 5{,}796 cards to share among 3 tables. How many cards does each table get?

Part 2: the short record

7. Make the short record for 8{,}472 \div 6.
8. Find 7{,}335 \div 5 with the short record. Which small digits do you write?
9. Find 9{,}744 \div 8.
10. In Magnhild’s short record for 6{,}184 \div 4, a small 2 is written in front of the 1. What does that small 2 mean? Finish the record.
11. A shoe store sets 3{,}975 pairs of shoes on 3 shoe racks, the same number on each. How many pairs go on each rack?
12. Find 8{,}595 \div 5. Then write 8{,}595 as 5 times a sum of thousands, hundreds, tens and ones, using the distributive property.

Part 3: first digits and zeros

13. Find 3{,}618 \div 9.
14. Find 1{,}535 \div 5.
15. Find 2{,}856 \div 7.
16. Find 4{,}820 \div 4.
17. Ariki puts 6{,}048 pairs of shoes on 6 shoe racks, the same number on each. How many pairs go on each rack?
18. Ariki says 2{,}135 \div 7 = 35. Show that he is wrong by multiplying back, and fix his answer.

Part 4: remainders and multiplying back

19. Find 3{,}467 \div 6. Check by multiplying back.
20. Find 8{,}123 \div 4. Check by multiplying back.
21. Is 5{,}290 \div 8 = 661 \text{ R } 2 right? Check by multiplying back.
22. Magnhild says 2{,}749 \div 3 = 915 \text{ R } 4. Her answer passes the multiply-back test. Why is it still wrong? Fix it.

Reasoning

23. A whole number divided by 9 gives a quotient of 452 and some remainder. Which numbers can the remainder be? What are the smallest and the largest the dividend can be? Explain.
24. Without dividing, explain how to tell whether a four-digit number divided by a one-digit number has a four-digit or a three-digit quotient.
25. Explain why multiplying the quotient by the divisor and adding the remainder gives back the dividend, using what the sharing did with every piece.