Dividing multi-digit decimals with the standard algorithm

When you reach for it

Odile is a carpenter. She has a length of dowel 3.15 meters long and wants pegs 0.45 meters long. How many pegs does the dowel make? The question is 3.15 \div 0.45, and this time the divisor has a decimal point.

Dividing by a whole number is already in hand from Dividing decimals: the long division runs exactly as it does for whole numbers, and the point in the quotient goes straight above the point in the dividend. For 8.4 \div 4, the 8 ones give 2 ones each and the 4 tenths give 1 tenth each, so 8.4 \div 4 = 2.1.

A decimal divisor is different. Long division finds each digit of the quotient by asking how many times the divisor fits into a whole number of some place, and “how many 0.45s fit into 3” is not a question one digit at a time can answer. The move in this lesson makes any decimal divisor whole without changing the quotient, then runs the long division you already know.

Reach for it whenever the divisor has a decimal point. It is taught over three days. Day 1 makes the divisor whole. Day 2 writes zeros in the dividend, to make room for that move and to carry the division past the last digit. Day 3 deals with quotients that run long or never end.

The words are the ones from whole-number division: in 3.15 \div 0.45 = 7, the dividend is 3.15, the divisor is 0.45 and the quotient is 7.

The procedure

Day 1: make the divisor a whole number

Measuring in smaller units. Count Odile’s dowel in hundredths of a meter. The dowel is 315 hundredths long and each peg is 45 hundredths. The number of pegs does not depend on the unit the lengths are measured in, so it is the number of 45s in 315:

Odile gets 7 pegs, and indeed 7 \times 0.45 = 3.15.

Why the quotient stays the same. A quotient is the number that, multiplied by the divisor, gives the dividend. Here 7 \times 0.45 = 3.15. Multiply both sides by 100: (7 \times 0.45) \times 100 = 3.15 \times 100. By the associative property of multiplication, the left side can be grouped as 7 \times (0.45 \times 100) = 7 \times 45, so 7 \times 45 = 315. The same 7 is the quotient of 315 \div 45. Nothing about the 7 or the 100 mattered: whenever a quotient times the divisor gives the dividend, the same quotient times 100 times the divisor gives 100 times the dividend. Multiplying the dividend and the divisor by the same number leaves the quotient unchanged. Written as fractions it is the familiar fact about Equivalent fractions, \displaystyle \frac{3.15}{0.45} = \frac{315}{45}.

Which power of ten. The aim is a whole-number divisor, so the divisor decides. A divisor with one decimal place is multiplied by 10, one with two places by 100, one with three by 1000. Multiplying by 10, 100 or 1000 moves every digit one, two or three places to the left, which is the same as moving the decimal point that many places to the right (Multiplying a decimal by ten, one hundred or one thousand). The dividend is multiplied by the same number, so its point moves the same number of places. The dividend’s own decimal places play no part in the choice, and after the move the dividend may still have a point. That is fine: it is then a decimal divided by a whole number, which is the case already in hand.

A dividend that keeps its point. Bram buys gasoline at 1.8 dollars a liter and the pump shows 61.92 dollars. He bought 61.92 \div 1.8 liters. The divisor has one decimal place, so both numbers are multiplied by 10:

Before dividing, estimate with Friendly numbers and deliberate over and under estimates, numbers near the real ones that divide easily: 60 \div 2 = 30, so expect about 30 liters. Now divide. The 6 is smaller than 18, so start with 61 tens: 3 each, since 3 \times 18 = 54, and 7 tens are left. Bring down the 9 to make 79 ones: 4 each, 4 \times 18 = 72, 7 left. Bring down the 2, which sits after the point, to make 72 tenths: 4 each, and nothing is left. The 4 just found counts tenths, so it goes after the point, and the quotient’s point sits straight above the dividend’s:

Bram bought 34.4 liters, close to the estimate of 30. Multiplying back, as in Multiplying multi-digit decimals with the standard algorithm, checks it exactly: 34.4 \times 1.8 = 61.92.

Day 2: writing zeros in the dividend

Zeros that make room. Take 6 \div 0.25. The divisor has two decimal places, so both numbers are multiplied by 100. But 6 has no decimal places for its point to move through. Write it as 6.00, which is allowed because 0 tenths and 0 hundredths add nothing. Now the point moves two places, and 6.00 \times 100 = 600:

and 24 \times 0.25 = 6. These zeros fill places the move needs.

Zeros that keep the division going. Take 4.2 \div 0.8, which becomes 42 \div 8. Whole-number long division stops at 5 \text{ R } 2, because no digits are left to bring down. But a remainder of 2 ones can be traded for 20 tenths, exactly the trade the bring-down step makes at every place. So write the dividend as 42.00 and keep going. Bring down a zero to make 20 tenths: 2 each, 2 \times 8 = 16, 4 left. Bring down another zero to make 40 hundredths: 5 each, and nothing is left.

Check: 5.25 \times 0.8 = 4.2. Write as many zeros as the division needs; none of them changes the dividend. The division ends when a subtraction leaves 0 and only zeros remain to bring down.

Stopping at 5 \text{ R } 2 would also mislead: that 2 counts ones of 42, but in the original numbers 5 \times 0.8 = 4, so what is really left over is 4.2 - 4 = 0.2.

Zeros in the quotient. As with whole numbers, once the quotient has begun, every bring-down puts one digit in the quotient, even when that digit is 0. A zero can sit right after the point, as in 18.54 \div 6 = 3.09, where 5 tenths are too few to give each of 6 groups one tenth (Example 6).

A quotient less than one. Take 0.36 \div 0.9, which becomes 3.6 \div 9. Three ones are too few for 9 groups to get one each, so the quotient has 0 ones, written as 0 before the point. Then 36 tenths give 4 tenths each:

and 0.4 \times 0.9 = 0.36. Whenever the dividend is smaller than the divisor the quotient is less than 1, and a 0 in the ones place says so.

Day 3: long quotients, stopping and rounding

Divisions that never end. Take 2 \div 0.3, which becomes 20 \div 3. Twenty ones give 6 each, 6 \times 3 = 18, and 2 are left. Bring down a zero: 20 tenths, 6 each, 2 left. Bring down a zero: 20 hundredths, and so on.

The same remainder with the same zero brought down makes the same step again, so the 6s go on forever: 2 \div 0.3 = 6.666\ldots. This is not bad luck. Every remainder is less than the divisor, so with divisor 3 the only possible remainders are 0, 1 and 2, and once the digits of the dividend run out, only zeros are brought down. Either a remainder of 0 turns up and the division ends, or some remainder comes back and from then on the digits repeat.

Stopping by rounding. In practice a quotient is given to the place the question needs: cents are hundredths of a dollar, millimeters are thousandths of a meter. Divide to one place past the one you need, then round using that extra digit. For 2 \div 0.3 to the nearest hundredth, divide to thousandths, 6.666, and round up because the thousandths digit is 6:

One extra digit is enough. Everything after the hundredths place is less than one hundredth, and the thousandths digit alone says whether it is less than half a hundredth (digits 0 to 4) or at least half (digits 5 to 9). The sign \approx, “is approximately equal to”, marks a rounded answer. Multiplying back no longer comes out exactly: 6.67 \times 0.3 = 2.001, a little over 2 because 6.67 was rounded up.

Checking with compatible numbers. A point moved one place too few or too many makes the answer ten times too big or too small, which a rough estimate such as 60 \div 2 for 61.92 \div 1.8 catches at once. Expect this too: dividing by a number less than 1 gives a quotient bigger than the dividend, since more than one small piece fits into each whole (3.15 \div 0.45 = 7), and dividing by a number more than 1 gives a smaller one (61.92 \div 1.8 = 34.4).

Worked examples

Examples 1 to 3 go with Day 1, 4 to 6 with Day 2, and 7 and 8 with Day 3. As in The standard long division algorithm, each table has one row per place of the scaled division: “have” is the number in play at that place after the bring-down, counted in that place’s units; “digit” is the quotient digit; “used” is the digit times the divisor; “left” is have minus used.

Boundaries

The divisor cannot be 0. Moving the point in 0.0 or 0.00 only ever gives 0, and no number times 0 makes a dividend that is not 0.

Both numbers are scaled by the same power of ten. Scaling the dividend alone multiplies the quotient by that power; scaling the divisor alone divides it by that power. A larger power than needed is also correct, only longer: 3.15 \div 0.45 = 3150 \div 450.

The divisor must be a decimal that ends. A divisor such as 0.333\ldots, whose digits never stop, cannot be made whole by any power of ten. Division by such a number is done with fractions instead.

A quotient may never end. Then it is rounded to the place the question needs, and the rounded quotient is approximate: multiplying back gives a number close to the dividend, not equal to it.

Remainders belong to whole-number division. After scaling, a remainder counts units of the scaled dividend, so it is 10, 100 or 1000 times what is really left over. With decimals the division is carried on to a decimal quotient instead. When a story needs a whole-number answer, such as full bottles or whole pegs, the decimal quotient is cut back to a whole number, as in Remainders and what they mean.

Where this goes next

Most of the time the division arrives inside a question: a price per kilogram, how many pieces a length makes. Deciding what to divide by what is Decimal problems in words. The scaling move itself is looked at on its own in Dividing by a decimal.

The long quotients of Day 3 lead to a deeper fact. Dividing a fraction’s numerator by its denominator turns the fraction into a decimal (Converting between fractions and decimals), and the argument about remainders coming back shows that every such decimal either ends or repeats. The bar that marks the repeating part is Terminating and repeating decimals.

Exercises

Day 1: a whole-number divisor

Divide, and check by multiplying.

1. 9.12 \div 0.4
2. 1.68 \div 0.07
3. 0.585 \div 0.45
4. 27.36 \div 3.6
5. 4.788 \div 0.126
6. Odile cuts a dowel 2.73 meters long into pieces 0.21 meters long. How many pieces does she get?
7. Rewrite each division with a whole-number divisor and the same quotient, then divide. (a) 5.4 \div 0.09 \qquad (b) 0.072 \div 1.2 \qquad (c) 300 \div 0.015
8. Bram pays 49.40 dollars for gasoline at 1.9 dollars a liter. How many liters did he buy?

Day 2: zeros in the dividend

9. 9 \div 0.36
10. 5.1 \div 0.4
11. 0.6 \div 0.016
12. 2.43 \div 0.6
13. 0.42 \div 0.8
14. 13 \div 0.104
15. On the camping trip Ravi fills water bottles that hold 0.75 liters from a carrier holding 6 liters. How many bottles does he fill?
16. Long division of 35 \div 4 stops at 8 \text{ R } 3. Carry it on to write 3.5 \div 0.4 as a decimal. Then say what the 3 left over stands for in 3.5 \div 0.4.

Day 3: rounding and estimating

17. 5 \div 0.7, to the nearest hundredth.
18. 2.9 \div 0.06, to the nearest tenth.
19. 0.83 \div 0.9, to the nearest hundredth.
20. Bram pays 12 dollars for 7.4 liters of gasoline. What is the price per liter, to the nearest cent?
21. Keziah and Ines pay 23 dollars for 1.5 kilograms of coffee for the camping trip. What does the coffee cost per kilogram, to the nearest cent?
22. Use compatible numbers, not long division, to choose the quotient of 2.596 \div 0.44 from 0.59, 5.9 and 59. Then check your choice by multiplying.

Reasoning

23. Explain why the decimal places of the divisor, and not those of the dividend, decide how far both points move. Use 4.5 \div 0.009 to show it, and find the quotient.
24. Without dividing, say whether 4.8 \div 0.6 is bigger or smaller than 4.8, and explain why. Then find it.
25. Explain, by looking at the remainders in the long division, why the decimal quotient of 1 \div 0.3 never ends. Which digit repeats?