The standard long division algorithm

When you reach for it

A shipping depot has 852 boxes to load onto 3 trucks, the same number on each truck. No times-table fact gives 852 \div 3, and handing the boxes out three at a time would take hundreds of rounds. Partial quotients would get there, by taking away 3 \times 200, then 3 \times 80, then 3 \times 4, but you have to choose the chunks yourself and add them up at the end.

The standard long division algorithm is partial quotients with no choosing. It works through the dividend one place at a time, from the biggest place down, and each step writes exactly one digit of the quotient, in its own place. Reach for it whenever a whole number is too big to divide in your head, by a divisor of one digit or two, and whenever you want the remainder as well as the quotient. With a one-digit divisor each step is a basic fact, as in Dividing by a single digit; with a two-digit divisor each step needs an estimate first.

The words: in 852 \div 3 = 284, the number being shared, 852, is the dividend; the number it is divided by, 3, is the divisor; the answer, 284, is the quotient. What is left when no more equal groups can be made is the remainder.

On paper the divisor goes outside a bracket, the dividend inside it, and the quotient on top. Each digit of the quotient is written straight above the last digit of the part of the dividend it came from, which is how the quotient keeps its place value.

The procedure

Part 1: one place at a time, and the bring-down step

Read 852 as 8 hundreds, 5 tens and 2 ones, and share the biggest pieces first.

Hundreds. Share 8 hundreds among 3 trucks. Each truck can get 2 hundreds, since 2 \times 3 = 6 and 3 \times 3 = 9 is more than 8. Write 2 above the 8, in the hundreds place. Multiply to see how many were handed out: 2 \times 3 = 6 hundreds. Subtract to see what is left: 8 - 6 = 2 hundreds.

Bring down. Two hundreds cannot be shared among three trucks as whole hundreds, so trade them: 2 hundreds are 20 tens, and with the 5 tens already there that makes 25 tens. Writing the 5 beside the 2 to make 25 is the bring-down step. It is that trade, done on paper.

Tens. Share 25 tens: each truck gets 8 tens, since 8 \times 3 = 24 and 9 \times 3 = 27 is too many. Write 8 above the 5. Subtract: 25 - 24 = 1 ten left. Bring down the 2: one ten and two ones are 12 ones.

Ones. Share 12 ones: 4 each, since 4 \times 3 = 12. Write 4 above the 2. Subtract: 12 - 12 = 0, and there are no digits left to bring down. So 852 \div 3 = 284.

Every place runs the same four moves: divide, multiply, subtract, bring down. Each one has a reason.

• Divide finds the most that each group can get at this place. The digit goes above that place because it counts units of that place: the 2 in 284 means 2 hundreds.
• Multiply finds how many were handed out in total.
• Subtract finds what is left. What is left must be less than the divisor. If 3 or more were left, every truck could have had one more, so the digit was too small. If the product is bigger than what you had, the digit was too big.
• Bring down trades what is left for ten times as many of the next smaller unit, and joins it to the digit already in that place.

Why the answer is right: the three steps handed out 600, 240 and 12 boxes, which together are all 852. By the distributive property,

So these are partial quotients after all, the largest possible one at each place.

A quotient digit is never more than 9. What is left after a subtraction is at most 2 when the divisor is 3, so after a bring-down the number being divided is at most 29, and 29 holds fewer than ten 3s. The same holds for any divisor.

Where the first digit goes. If the first digit of the dividend is smaller than the divisor, no group gets one of that place. Start with the first two digits instead (with a two-digit divisor, take as many digits as it needs to reach a number at least as big as the divisor). Each digit skipped at the start is one digit fewer in the quotient (Example 2).

Checking. Multiplication undoes division (Multiplication and division undo each other), so multiply back: 284 \times 3 = 852.

Part 2: zeros in the quotient

Divide 624 by 3. Hundreds: 6 hundreds give 2 each, 2 \times 3 = 6, and 0 are left. Bring down the 2: there are now 2 tens. Tens: 2 tens shared among 3 gives each group no tens at all. The digit is 0, and it is written, above the 2. Then 0 \times 3 = 0, and 2 - 0 = 2 tens are still left. Bring down the 4: 2 tens and 4 ones are 24 ones. Ones: 8 each, since 8 \times 3 = 24, and nothing is left. So 624 \div 3 = 208.

The 0 holds the tens place. Leave it out and the 2 and the 8 slide together into 28, and 28 \times 3 = 84, nowhere near 624. The rule that prevents this: once the first digit of the quotient is written, every bring-down puts one more digit into the quotient, even when that digit is 0. So the quotient has a digit above every digit of the dividend from the first one used to the last.

An estimate catches a lost zero before you even multiply. 624 \div 3 is about 600 \div 3 = 200, so the answer is in the hundreds and has three digits. A zero can also come last, when the final number brought down is smaller than the divisor (Example 4).

Part 3: two-digit divisors and estimating each digit

Divide 896 by 32. The first digit, 8, is smaller than 32, so start with 89 tens. First estimate the whole answer with Friendly numbers and deliberate over and under estimates: 896 \div 32 is about 900 \div 30 = 30, so the quotient is about 30, with two digits.

Tens. How many 32s are in 89? That is not a basic fact, so estimate. Round the divisor to 30 and think of 89 as about 90: 9 tens divided by 3 tens is 3. Try 3: 3 \times 32 = 96, which is more than 89. Too big, so try 2: 2 \times 32 = 64, and 89 - 64 = 25, which is less than 32. The digit is 2.

Ones. Bring down the 6 to make 256. Estimate again: 256 \div 30 is about 25 \div 3, roughly 8. Try 8: 8 \times 32 = 256, and 256 - 256 = 0. So 896 \div 32 = 28, and 28 \times 32 = 896.

The four moves are the same as in Part 1. The only new thing is that the divide step is now a guess followed by two tests, and the tests are the ones from Part 1:

• the product is bigger than the number being divided: the guess is too big; lower it by one;
• what is left is as big as the divisor or bigger: the guess is too small; raise it by one.

The direction of the rounding tells you which way a guess tends to miss. Here 32 was rounded down to 30, and dividing by a smaller number gives a bigger answer, so the first guess came out too big. A divisor rounded up, such as 47 to 50, tends to give guesses that are too small (Example 6). Either way the tests catch it. Choosing trial multiples of a larger divisor is practiced further in Dividing by a two-digit number.

Part 4: remainders and checking

Divide 437 by 6, say 437 boxes packed onto pallets of 6. The 4 is smaller than 6, so start with 43 tens: 7 each, 7 \times 6 = 42, 1 left. Bring down the 7 to make 17: 2 each, 2 \times 6 = 12, 5 left. No digits are left to bring down, and 5 is less than 6, so no further group can be made. The 5 is the remainder, written

so the depot fills 72 pallets and has 5 boxes over.

Checking a division with a remainder needs two things:

and the remainder must be less than the divisor. Here 72 \times 6 + 5 = 432 + 5 = 437, and 5 < 6. Both conditions matter: 71 \times 6 + 11 is also 437, but 11 is more than 6, so 71 is not the quotient. With both conditions the quotient and remainder are the only possible pair, a fact stated in general as the division algorithm.

The check catches a wrong digit, a lost zero and a missed bring-down alike (Example 8). What a remainder means in a story, whether to round up, drop it or share it, is the subject of Remainders and what they mean.

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. In each table, one row is one place, and the columns follow the four moves: “have” is the number in play at that place after the bring-down, counted in that place’s units; “digit” is the quotient digit (divide); “used” is the digit times the divisor (multiply); and “left” is have minus used (subtract).

Boundaries

The divisor cannot be 0. No number of groups of 0 makes 852, so 852 \div 0 has no answer, and the first divide step has nothing to find.

The remainder is always less than the divisor, at every step and at the end. A leftover as big as the divisor means the digit above it was too small.

Each quotient digit is a single digit, 0 to 9. If an estimate says 10, take 9: a leftover less than the divisor, with one digit brought down, always holds fewer than ten divisors.

A dividend smaller than the divisor gives quotient 0 and leaves the whole dividend as the remainder: 5 \div 8 = 0 \text{ R } 5.

The method works for any divisor. A three-digit divisor needs no new idea, but the estimate step gets harder, and it is usually easier to think in multiples, as in Dividing by a two-digit number.

Whole numbers only. Here the division stops at the ones place and leaves a remainder. Carrying on past the ones place, with a decimal point, gives a decimal quotient instead; that is Dividing decimals and Dividing multi-digit decimals with the standard algorithm.

Where this goes next

The same divide, multiply, subtract, bring down is the engine of every later division. Taken past the ones place it gives decimal quotients, first when the dividend has a decimal point (Dividing decimals) and then for any decimals (Dividing multi-digit decimals with the standard algorithm). Run on a fraction’s numerator and denominator it turns the fraction into a decimal (Converting between fractions and decimals), and there the remainders that come back again are what make some decimals repeat.

The check from Part 4, quotient times divisor plus remainder equals dividend, with the remainder smaller than the divisor, is The division algorithm, the fact behind remainders, divisibility and the greatest common factor. And longer two-digit divisions, where choosing the trial multiple is most of the work, are Dividing by a two-digit number.

Exercises

Part 1: one-digit divisors

Divide, and check by multiplying.

1. 96 \div 4
2. 738 \div 3
3. 4{,}165 \div 5
4. 9{,}184 \div 7
5. Babajide shares 2{,}592 boxes equally among 8 trucks. How many boxes go on each truck?

Part 2: zeros in the quotient

6. 618 \div 6
7. 2{,}840 \div 4
8. 5{,}445 \div 5
9. 8{,}016 \div 8
10. Signe works out 7{,}231 \div 7 and gets 133. Show that she is wrong without dividing, then find the right quotient.

Part 3: two-digit divisors

11. 672 \div 21
12. 1{,}564 \div 34
13. 3{,}096 \div 43
14. 8{,}265 \div 15
15. 5{,}488 \div 56
16. 6{,}496 \div 32
17. Oona has 4{,}368 boxes and pallets that hold 52. How many pallets does she fill?

Part 4: remainders and checking

Divide, give the remainder, and check.

18. 875 \div 4
19. 5{,}003 \div 6
20. 2{,}917 \div 23
21. 4{,}000 \div 64
22. Ahmad says 3{,}145 \div 8 = 392 \text{ R } 9, and points out that 392 \times 8 + 9 = 3{,}145. Is his answer right? If not, correct it.
23. Teodor has 1{,}250 boxes to move, and each truck carries 36. How many trucks are full, how many boxes are left over, and how many trucks does he need to move every box?

Reasoning

24. Mina is dividing by 7. At one step her subtraction leaves 9. What went wrong, and how does she fix it?
25. Without dividing, explain how many digits the quotient of 3{,}412 \div 43 has. Then divide.
26. Liesel divides 745 by 5. After the hundreds step 2 is left, and she brings down the 4 to make 24. Explain what the 24 counts, and why the next quotient digit goes above the 4 and not above the 7. Then finish the division.