Grade 5 · 5.NBT.B.6 · 6.NS.B.2

Dividing by a two-digit number

When the divisor has two digits, no times table tells you each digit of the quotient. How do you find it?

You need

The standard long division algorithmGrade 5

Where you are 4 in Long division

Number9

Arithmetic5

Multiplication and division16

Long division4

Dividing by a single digit Partial quotients The standard long division algorithm

Dividing by a two-digit number

When you reach for it

Akira works at a bookstore. A delivery of 1{,}768 books has come in, and each shipping box holds 34 books. How many boxes will she fill?

That is 1{,}768 \div 34. The layout is the one from the standard long division algorithm: divide, multiply, subtract, bring down, one place at a time. What changes with a divisor of two digits is the divide step. “How many 3s are in 17?” is a basic fact. “How many 34s are in 176?” is not. You have to make a good guess and then test it.

Reach for this skill whenever the divisor is a two-digit number: books per box, seats per row, dollars per ticket. Partial quotients would also work, but it lets you take away any chunk you like, so it can take many rounds. Long division takes the biggest chunk at each place, so it finishes in one row per digit.

This lesson takes four days. Day 1 is a quotient with one digit. Day 2 is quotients with two or three digits. Day 3 is what to do when a guess is wrong. Day 4 is longer dividends, zeros in the quotient, and what a remainder means in a story.

The procedure

Part 1: guessing a digit by rounding the divisor

Start with a small one. Akira has 238 books and boxes that hold 34. How many boxes?

Where the digit goes. Look at the dividend from the left. 2 hundreds cannot fill a box of 34 as hundreds. 23 tens is still less than 34. So the first place that can be divided is the ones: all 238 ones. The quotient has just one digit, and it goes above the 8.

The guess. How many 34s are in 238? Round the divisor to the nearest ten: 34 is close to 30. Now ask how many 30s are in 238. Thirty is 3 tens, and 238 is 23 tens and 8 ones, so cover the ones digit and think 23 \div 3. The fact 7 \times 3 = 21 fits, and 8 \times 3 = 24 is too many. The guess is 7.

The test. Multiply: 7 \times 34 = 238. Subtract: 238 - 238 = 0. The two tests from the long division lesson both pass:

the product, 238, is not bigger than what we had, so the guess is not too big;
what is left, 0, is less than 34, so the guess is not too small.

Akira fills 7 boxes.

Figure 1238 \div 34 = 7. Neither 2 nor 23 holds a 34, so the only digit of the quotient sits above the ones place.

Why covering the ones digit works. Counting 30s in 238 is the same as counting 3 tens in 23 tens. The 8 loose ones cannot make another 30, so they do not change the answer. That is why the guess only ever needs a basic fact.

Why the test is needed. The guess was made with 30, not with 34. The real divisor is a little bigger, so the guess might be one too many, or (when the divisor was rounded up) one too few. The multiply and the subtract are what decide. On Day 1 the guesses happen to pass. Day 3 is about the ones that do not.

Checking. Multiplication undoes division, so multiply back: 7 \times 34 = 238. With a remainder, check that quotient times divisor plus remainder gives the dividend, and that the remainder is less than the divisor.

Part 2: two- and three-digit quotients

Now the real delivery: 1{,}768 \div 34.

Estimate the whole answer first. Use Friendly numbers and deliberate over and under estimates: 1{,}768 is near 1{,}800, and 34 is near 30, so the answer is near 1{,}800 \div 30 = 60. Expect about 50 or 60, a two-digit quotient.

Where the first digit goes. 1 thousand and 17 hundreds are both less than 34. The first number that holds a 34 is 176 tens. So the first digit goes above the 6, in the tens place, and the quotient has two digits, as the estimate said.

Tens. How many 34s are in 176? Round 34 to 30, cover the last digit: 17 \div 3 gives 5. Multiply: 5 \times 34 = 170. Subtract: 176 - 170 = 6, which is less than 34. The tens digit is 5. Those 5 tens of boxes hold 50 \times 34 = 1{,}700 books.

Bring down. The 6 tens left trade for 60 ones, and with the 8 ones that makes 68.

Ones. How many 34s are in 68? Cover the last digit: 6 \div 3 = 2. Multiply: 2 \times 34 = 68. Subtract: 0. The ones digit is 2.

Akira fills 52 boxes. Check: 52 \times 34 = 1{,}700 + 68 = 1{,}768.

Figure 21{,}768 \div 34 = 52. Each step makes a fresh estimate, and each estimate is tested by the multiply and the subtract.

Every place gets its own guess. The rounded divisor stays the same (30 every time), but what you divide it into changes after each bring-down. A longer quotient is just more of the same step.

A bound on each digit. Each digit is between 0 and 9. After a subtraction, what is left is less than the divisor. Bringing down one digit makes it at most ten times that, plus a little, and that still holds fewer than ten divisors. So if a guess ever says 10 or more, try 9.

Part 3: when a guess is too big or too small

Hadiza unpacks 4{,}503 books onto shelves that hold 79 each. Divide 4{,}503 by 79.

Start. 45 hundreds is less than 79, so start with 450 tens. Round 79 to 80. Estimate: 4{,}800 \div 80 = 60, so expect a two-digit answer a bit under 60.

Tens. How many 80s are in 450? Cover the last digit: 45 \div 8 gives 5. Multiply: 5 \times 79 = 395. Subtract: 450 - 395 = 55, less than 79. The digit is 5.

Ones. Bring down the 3: 553. How many 80s are in 553? 55 \div 8 gives 6. Multiply: 6 \times 79 = 474. Subtract: 553 - 474 = 79.

What is left is 79, and that is not less than 79. A whole shelf’s worth of books is still sitting there, so one more shelf can be filled. The guess 6 is too small. Raise it to 7: 7 \times 79 = 553, and 553 - 553 = 0. The digit is 7.

Hadiza fills 57 shelves. Check: 57 \times 79 = 57 \times 80 - 57 = 4{,}560 - 57 = 4{,}503.

Figure 34{,}503 \div 79 = 57. Only the digits that passed both tests are written; the first guess for the ones digit, 6, was too small.

The two tests each point one way:

What you see	What it means	What to do
the product is bigger than the number you have	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

Which way a guess misses. Here the guess is how many of the rounded divisor fit. A bigger divisor never fits more times into the same number. So:

if you round the divisor up (like 79 to 80), you are dividing by something too big, and the guess can only be right or too small, never too big;
if you round the divisor down (like 34 to 30), the guess can only be right or too big, never too small.

This tells you which test to watch. It also means one fix may not be enough. A divisor whose ones digit is near 5, such as 15 or 25, is far from both tens, so a guess can miss by two or more. Raise or lower one step at a time, and test each time (Example 6).

Another way to guess. When a guess keeps missing, write out the multiples of the divisor: 79, 158, 237, 316, 395, 474, 553, 632, 711. Then each step is “find the biggest multiple that fits”, and nothing has to be fixed. It costs a few multiplications at the start, and it pays off when the quotient is long.

Part 4: longer dividends, zeros, and remainders

Octavio packs 9{,}450 books into boxes of 27. Divide 9{,}450 by 27.

Start. 9 thousands is less than 27, so start with 94 hundreds. Estimate: 9{,}000 \div 30 = 300, so the quotient has three digits, in the hundreds.

Hundreds. Round 27 to 30. 94 \div 30: cover the last digit, 9 \div 3 = 3. Multiply: 3 \times 27 = 81. Subtract: 94 - 81 = 13. The digit is 3.

Tens. Bring down the 5: 135. 13 \div 3 gives 4. Multiply: 4 \times 27 = 108. Subtract: 135 - 108 = 27. That is not less than 27, so 4 is too small (rounding 27 up made that possible). Try 5: 5 \times 27 = 135, and 0 is left. The digit is 5.

Ones. Bring down the 0: there are 0 ones. No box gets any, so the digit is 0, and it is written. Then 0 \times 27 = 0 and 0 - 0 = 0.

Octavio fills 350 boxes. Check: 350 \times 27 = 9{,}450.

Figure 49{,}450 \div 27 = 350. The last digit brought down gives 0 ones, so the ones digit of the quotient is 0.

Zeros in the quotient. Once the first digit is written, every bring-down puts one more digit in the quotient. When the number in play after a bring-down is smaller than the divisor, that digit is 0. Leave it out and the answer here would read 35, and 35 \times 27 = 945, ten times too small. The estimate of 300 catches that at a glance.

Remainders in a story. When the last subtraction leaves something less than the divisor but not 0, that is the remainder. What to do with it depends on the question, not on the division:

“How many full boxes?” Drop the remainder; the quotient is the answer.
“How many boxes to hold all the books?” The leftover books need one more box, so add one.
“How many books are left over?” The remainder itself is the answer.

Remainders and what they mean looks at these choices more closely.

Checking by multiplying is the last step of every division:

\text{quotient} \times \text{divisor} + \text{remainder} = \text{dividend}

and the remainder must be less than the divisor. Both must hold. The check catches a wrong digit, a lost zero, and a subtraction slip alike (Example 8).

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 the tables, one row is one place. “Have” is the number in play at that place after the bring-down, “trial” is the guess, “used” is the trial times the divisor, and “left” is have minus used.

Example 1

A one-digit quotient with a remainder

Hadiza has 461 books and shelves that hold 72 books each. How many shelves can she fill, and how many books are left over?

Solution

4 and 46 are both less than 72, so start with all 461 ones. The quotient has one digit, above the 1.

Round 72 to 70. How many 70s are in 461? Cover the last digit: 46 \div 7. Since 6 \times 7 = 42 and 7 \times 7 = 49, the guess is 6.

Multiply: 6 \times 72 = 432, which is not more than 461. Subtract: 461 - 432 = 29, which is less than 72. Both tests pass. There are no more digits to bring down.

461 \div 72 = 6 \text{ R } 29

Hadiza fills 6 shelves and has 29 books left over.

Check: 6 \times 72 + 29 = 432 + 29 = 461, and 29 < 72.

Example 2

How many digits?

Divide (a) 847 \div 93 \qquad (b) 847 \div 63. Before dividing, say how many digits each quotient has.

Solution

(a) 84 is less than 93, so the first number that holds a 93 is all 847 ones. The quotient has one digit. Round 93 to 90: 84 \div 9 gives 9. Multiply: 9 \times 93 = 837. Subtract: 847 - 837 = 10, less than 93. So 847 \div 93 = 9 \text{ R } 10.

Check: 9 \times 93 + 10 = 837 + 10 = 847.

(b) This time 84 tens is at least 63, so the first digit goes above the 4, in the tens place, and the quotient has two digits. Round 63 to 60.

Have	Trial	Used	Left
84 tens	1	63	21
217 ones	3	189	28

The second trial came from 21 \div 6, which gives 3. So 847 \div 63 = 13 \text{ R } 28.

Check: 13 \times 63 + 28 = 819 + 28 = 847.

The same dividend gave quotients of different lengths. What decides it is how many digits of the dividend it takes to reach the divisor.

Example 3

A three-digit quotient

Octavio has 7{,}222 books to pack, 23 to a box. How many boxes?

Solution

7 < 23 but 72 \ge 23, so start with 72 hundreds. The quotient has three digits. Estimate: 7{,}000 \div 20 = 350, but 23 is bigger than 20, so the answer is a bit less, around 300.

Round 23 to 20, so each guess comes from dividing by 2 after covering the last digit.

Have	Trial	Used	Left
72 hundreds	3	69	3
32 tens	1	23	9
92 ones	4	92	0

The trials came from 7 \div 2, 3 \div 2 and 9 \div 2. Every product fit, and every “left” is less than 23, so each trial passed. Octavio fills 314 boxes.

Check: 300 \times 23 = 6{,}900 and 14 \times 23 = 322, so 314 \times 23 = 7{,}222.

Example 4

Estimate first, then divide

Divide 5{,}467 by 81. Estimate first, and use the estimate to judge the answer.

Solution

Estimate with compatible numbers: 5{,}600 \div 80 = 70. So expect a two-digit answer near 70.

54 hundreds is less than 81, so start with 546 tens. Round 81 to 80.

Have	Trial	Used	Left
546 tens	6	486	60
607 ones	7	567	40

The trials came from 54 \div 8 and 60 \div 8. So 5{,}467 \div 81 = 67 \text{ R } 40, which is close to the estimate of 70.

Check: 67 \times 81 = 5{,}427, and 5{,}427 + 40 = 5{,}467, with 40 < 81.

Example 5

A guess that is too big

Akira divides 3{,}687 books among boxes of 62. How many full boxes, and how many books are left?

Solution

36 hundreds is less than 62, so start with 368 tens. Round 62 to 60. Rounding down means a guess may be too big, so watch the product.

Tens. 36 \div 6 = 6. Multiply: 6 \times 62 = 372. That is more than 368: 6 tens of boxes would need more books than there are. Too big. Lower it to 5: 5 \times 62 = 310, and 368 - 310 = 58, which is less than 62. The digit is 5.

Ones. Bring down the 7: 587. 58 \div 6 gives 9. Multiply: 9 \times 62 = 558. Subtract: 587 - 558 = 29, less than 62. The digit is 9.

3{,}687 \div 62 = 59 \text{ R } 29

Akira fills 59 boxes and has 29 books left.

Check: 59 \times 62 = 3{,}658, and 3{,}658 + 29 = 3{,}687.

Example 6

A guess that misses by two

Octavio packs 945 books into boxes of 15. How many boxes?

Solution

9 < 15, so start with 94 tens. 15 rounds to 20, a big jump up, so guesses may be too small, maybe by more than one.

Tens. 94 \div 20: 9 \div 2 gives 4. Multiply: 4 \times 15 = 60. Subtract: 94 - 60 = 34. That is more than 15, so 4 is too small. Try 5: 5 \times 15 = 75, and 94 - 75 = 19, still more than 15. Try 6: 6 \times 15 = 90, and 94 - 90 = 4. The digit is 6.

Ones. Bring down the 5: 45. 4 \div 2 = 2. Multiply: 2 \times 15 = 30, and 45 - 30 = 15, which is not less than 15. Too small again. Try 3: 3 \times 15 = 45, and 0 is left. The digit is 3.

Octavio fills 63 boxes. Check: 63 \times 15 = 900 + 45 = 945.

The guesses kept missing because 15 is far from 20. The list of multiples avoids that: 15, 30, 45, 60, 75, 90, 105. The biggest that fits in 94 is 90 = 6 \times 15, and the biggest that fits in 45 is 45 = 3 \times 15. One look each, no fixing.

Example 7

A zero in the middle and a remainder in a story

Hadiza has 4{,}053 books for shelves that hold 38 each. (a) How many shelves can she fill completely? (b) How many shelves does she need to hold every book?

Solution

Start with 40 hundreds, since 4 < 38. Estimate: 4{,}000 \div 40 = 100, so the quotient has three digits and begins with 1. Round 38 to 40.

Have	Trial	Used	Left
40 hundreds	1	38	2
25 tens	0	0	25
253 ones	6	228	25

After the hundreds, 2 is left, and bringing down the 5 gives only 25 tens. That is less than 38, so the tens digit is 0, and the 25 tens are carried on to become 250 ones. With the 3 that makes 253, and 25 \div 4 gives 6.

4{,}053 \div 38 = 106 \text{ R } 25

(a) She can fill 106 shelves completely. The question asks for full shelves, so the remainder is dropped.

(b) The 25 leftover books still need a shelf, so she needs 106 + 1 = 107 shelves.

Check: 106 \times 38 = 4{,}028, and 4{,}028 + 25 = 4{,}053, with 25 < 38.

Example 8

Using the check to catch a lost zero

Akira works out 8{,}534 \div 42 and writes 23 \text{ R } 8. Check her answer, and correct it if it is wrong.

Solution

Multiply back: 23 \times 42 + 8 = 966 + 8 = 974. That is nowhere near 8{,}534, so the answer is wrong. An estimate shows how: 8{,}000 \div 40 = 200, so the quotient should have three digits.

Divide again, rounding 42 to 40.

Have	Trial	Used	Left
85 hundreds	2	84	1
13 tens	0	0	13
134 ones	3	126	8

The 13 tens are fewer than 42, so the tens digit is 0. Akira skipped it and wrote 23 instead of 203.

8{,}534 \div 42 = 203 \text{ R } 8

Check: 203 \times 42 = 8{,}526, and 8{,}526 + 8 = 8{,}534, with 8 < 42.

Boundaries

The divisor cannot be 0. No number of groups of 0 makes 238, so dividing by 0 has no answer.

Rounding the divisor is only for the guess. The multiply and the subtract always use the real divisor. A guess made with 30 and tested with 30 would give the answer to a different question.

Each quotient digit is 0 to 9. If a guess says 10 or more, take 9 and test it.

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

A dividend smaller than the divisor gives quotient 0 with the whole dividend left over: 25 \div 38 = 0 \text{ R } 25.

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

Bigger divisors work the same way. A three-digit divisor uses the same four moves; the guess is made by rounding to the nearest hundred, and the tests are unchanged.

Where this goes next

Dividing by a two-digit number is the last new piece of whole-number long division. From here the same steps carry past the ones place into decimals (Dividing multi-digit decimals with the standard algorithm), where a remainder becomes more digits after the decimal point instead of a leftover.

The check you used every day, quotient times divisor plus remainder equals dividend with the remainder smaller than the divisor, is The division algorithm. That fact sits behind divisibility, factors and the greatest common factor. And deciding what a remainder means, drop it, add one, or report it, is the subject of Remainders and what they mean.

Exercises

Part 1: one-digit quotients

Divide, and check by multiplying.

156 \div 52
294 \div 42
415 \div 83
380 \div 61
Akira has 434 books and boxes that hold 53. How many boxes can she fill, and how many books are left?
In 627 \div 74, explain why the quotient has only one digit and which digit of 627 it is written above. Then divide.

Part 2: two- and three-digit quotients

1{,}118 \div 43
2{,}852 \div 31
5{,}828 \div 62
8{,}673 \div 21
9{,}856 \div 44
Hadiza packs 3{,}240 books into boxes of 72. How many boxes does she fill?

Part 3: adjusting the guess

For each, say whether any guess was too big or too small, and fix it.

1{,}156 \div 17
2{,}548 \div 52
6{,}120 \div 85
4{,}209 \div 61
Octavio packs 1{,}725 books into boxes of 25. How many boxes? Then give a quicker way to make each guess.
Akira rounds 46 down to 40 and tries 7 for the first digit of 2{,}984 \div 46. Is 7 right? Finish the division.

Part 4: zeros, remainders, and checking

6{,}380 \div 29
7{,}548 \div 37
5{,}000 \div 48
Hadiza has 2{,}500 books, and each box holds 36. How many boxes does she need to pack every book?
Octavio has 3{,}170 books, and each shelf holds 45. How many shelves can he fill completely, and how many books are left?

Reasoning

Hadiza says: “When I round the divisor up, my guess is never too big.” Explain why she is right, and say what is true when the divisor is rounded down.
Octavio works out 2{,}342 \div 34 and gets 68 \text{ R } 12. Check his answer by multiplying. If it is wrong, find the right answer.

Answers

3. 156 ones; 15 \div 5 = 3; 3 \times 52 = 156, 0 left.
7. 29 \div 4 gives 7; 7 \times 42 = 294, 0 left.
5. 41 \div 8 gives 5; 5 \times 83 = 415, 0 left.
6 \text{ R } 14. 38 \div 6 gives 6; 6 \times 61 = 366; 380 - 366 = 14. Check: 366 + 14 = 380.
8 boxes, 10 books left. 43 \div 5 gives 8; 8 \times 53 = 424; 434 - 424 = 10, less than 53.
8 \text{ R } 35. 6 and 62 are both less than 74, so the first number that holds a 74 is all 627 ones: one digit, above the 7. 62 \div 7 gives 8; 8 \times 74 = 592; 627 - 592 = 35. Check: 592 + 35 = 627.
26. 111 tens: 2, 86, 25 left; 258: 25 \div 4 gives 6, 258, 0 left. Check: 26 \times 43 = 1{,}118.
92. 285 tens: 9, 279, 6 left; 62: 2, 0 left. Check: 92 \times 31 = 2{,}852.
94. 582 tens: 9, 558, 24 left; 248: 4, 0 left. Check: 94 \times 62 = 5{,}828.
413. 86 hundreds: 4, 84, 2 left; 27: 1, 6 left; 63: 3. Check: 413 \times 21 = 8{,}673.
224. 98: 2, 88, 10 left; 105: 2, 88, 17 left; 176: 4, 0 left. Check: 224 \times 44 = 9{,}856.
45 boxes. 324 tens: 32 \div 7 gives 4, 288, 36 left; 360: 5, 0 left. Check: 45 \times 72 = 3{,}240.
68; every miss was too small (rounding 17 up to 20). 115: guess 5 leaves 30, too small, so 6, 102, 13 left. 136: guess 6 leaves 34, then 7 leaves 17, both too small, so 8, 136, 0 left. Check: 68 \times 17 = 1{,}156.
49; one guess too big (rounding 52 down to 50). 254: guess 5 gives 260, too big, so 4, 208, 46 left; 468: 9, 0 left. Check: 49 \times 52 = 2{,}548.
72; two guesses too small (rounding 85 up to 90). 612: guess 6 leaves 102, too small, so 7, 595, 17 left; 170: guess 1 leaves 85, too small, so 2, 0 left. Check: 72 \times 85 = 6{,}120.
69; one guess too big. 420: guess 7 gives 427, too big, so 6, 366, 54 left; 549: 9, 0 left. Check: 69 \times 61 = 4{,}209.
69 boxes. With 25 rounded to 30: 172 gives guess 5, leaving 47, too small, so 6, 150, 22 left; 225 gives guess 7, then 8, both too small, so 9, 0 left. Quicker: 25 is a quarter of 100, so its multiples 25, 50, 75, 100, \dots are easy to list, and 150 = 6 \times 25 and 225 = 9 \times 25 can be read off. Check: 69 \times 25 = 1{,}725.
No, 7 is too big: 7 \times 46 = 322, more than 298. Use 6: 276, 22 left; bring down the 4: 224. 22 \div 4 gives 5, but 5 \times 46 = 230 is too big, so the digit is 4: 184, 40 left. So 2{,}984 \div 46 = 64 \text{ R } 40. Check: 64 \times 46 = 2{,}944, and 2{,}944 + 40 = 2{,}984.
220. 63 hundreds: 2, 58, 5 left; 58: guess 1 leaves 29, too small, so 2, 0 left; 0 ones: digit 0. Check: 220 \times 29 = 6{,}380.
204. 75 hundreds: guess 1 leaves 38, too small, so 2, 74, 1 left; 14 tens is less than 37: digit 0; 148: guess 3 leaves 37, too small, so 4, 0 left. Check: 204 \times 37 = 7{,}548.
104 \text{ R } 8. 50: 1, 2 left; 20: digit 0; 200: 4, 192, 8 left. Check: 104 \times 48 + 8 = 4{,}992 + 8 = 5{,}000.
70 boxes. 2{,}500 \div 36 = 69 \text{ R } 16: 250: 6, 216, 34 left; 340: guess 8 leaves 52, too small, so 9, 324, 16 left. The 16 leftover books need one more box: 69 + 1 = 70. Check: 69 \times 36 + 16 = 2{,}484 + 16 = 2{,}500.
70 full shelves, 20 books left. 3{,}170 \div 45 = 70 \text{ R } 20: 317: guess 6 leaves 47, too small, so 7, 315, 2 left; 20 ones: digit 0, 20 left. Check: 70 \times 45 + 20 = 3{,}150 + 20 = 3{,}170.
A bigger divisor never fits more times into the same number. The rounded-up divisor is at least as big as the real one, so the number of rounded divisors that fit is never more than the number of real divisors that fit: the guess is right or too small. Rounded down, the rounded divisor is smaller than the real one and fits at least as many times, so the guess is right or too big, never too small.
Wrong: 68 \times 34 + 12 = 2{,}312 + 12 = 2{,}324, not 2{,}342. The right answer is 68 \text{ R } 30. 234 tens: guess 7 gives 238, too big, so 6, 204, 30 left; 302: 30 \div 3 = 10, so try 9: 306, too big, so 8, 272, 30 left. Check: 2{,}312 + 30 = 2{,}342, and 30 < 34. His quotient was right; the slip was in the last subtraction.