← The remainder is always less than the divisor · Divisibility and Remainder Reasoning

The remainder is always less than the divisor · 10 practice problems

4.NBT.B.6

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: box = 7 (then 377 / 14 = 26 remainder 13)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$37\square \div 14$

Show solution

1 · Understandwhat's really being asked

In the division 37box / 14, the box is a single digit 0-9 making 37box a number between 370 and 379. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 37box, where box is one digit from 0 to 9
The divisor is 14
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 13
The dividend ranges only from 370 to 379

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 14 can be at most 13, so we look for a dividend in the range 370-379 that leaves the biggest remainder. Finding the nearest multiple of 14 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 14 is 13; we hope to hit exactly remainder 13.

\text{remainder} < 14 \Rightarrow \text{max remainder} = 13

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 14 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 14 near 370. 14 x 26 = 364. Adding the leftover 13 gives 364 + 13 = 377, which is between 370 and 379.

14 \times 26 = 364,\quad 364 + 13 = 377

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 377 = 37box with box = 7, choose box = 7. Check: 377 / 14 = 26 remainder 13, the largest remainder available.

377 \div 14 = 26 \text{ R } 13

The dividend 377 sits in the allowed 370-379 window, so the digit is 7.

Answer: box = 7 (then 377 / 14 = 26 remainder 13)

4 · Reviewdoes it hold up?

Remainder 13 is less than the divisor 14, which is required. The dividend 377 is within 370-379, so the digit 7 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 7 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 370 through 379 by 14 and list the remainders; the largest, 13, occurs at 377, so box = 7.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 14 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 2 easy answer: box = 4 (then 454 / 13 = 34 remainder 12)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$45\square \div 13$

Show solution

1 · Understandwhat's really being asked

In the division 45box / 13, the box is a single digit 0-9 making 45box a number between 450 and 459. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 45box, where box is one digit from 0 to 9
The divisor is 13
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 12
The dividend ranges only from 450 to 459

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 13 can be at most 12, so we look for a dividend in the range 450-459 that leaves the biggest remainder. Finding the nearest multiple of 13 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 13 is 12; we hope to hit exactly remainder 12.

\text{remainder} < 13 \Rightarrow \text{max remainder} = 12

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 13 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 13 near 450. 13 x 34 = 442. Adding the leftover 12 gives 442 + 12 = 454, which is between 450 and 459.

13 \times 34 = 442,\quad 442 + 12 = 454

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 454 = 45box with box = 4, choose box = 4. Check: 454 / 13 = 34 remainder 12, the largest remainder available.

454 \div 13 = 34 \text{ R } 12

The dividend 454 sits in the allowed 450-459 window, so the digit is 4.

Answer: box = 4 (then 454 / 13 = 34 remainder 12)

4 · Reviewdoes it hold up?

Remainder 12 is less than the divisor 13, which is required. The dividend 454 is within 450-459, so the digit 4 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 4 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 450 through 459 by 13 and list the remainders; the largest, 12, occurs at 454, so box = 4.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 13 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 3 easy answer: box = 3 (then 543 / 16 = 33 remainder 15)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$54\square \div 16$

Show solution

1 · Understandwhat's really being asked

In the division 54box / 16, the box is a single digit 0-9 making 54box a number between 540 and 549. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 54box, where box is one digit from 0 to 9
The divisor is 16
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 15
The dividend ranges only from 540 to 549

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 16 can be at most 15, so we look for a dividend in the range 540-549 that leaves the biggest remainder. Finding the nearest multiple of 16 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 16 is 15; we hope to hit exactly remainder 15.

\text{remainder} < 16 \Rightarrow \text{max remainder} = 15

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 16 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 16 near 540. 16 x 33 = 528. Adding the leftover 15 gives 528 + 15 = 543, which is between 540 and 549.

16 \times 33 = 528,\quad 528 + 15 = 543

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 543 = 54box with box = 3, choose box = 3. Check: 543 / 16 = 33 remainder 15, the largest remainder available.

543 \div 16 = 33 \text{ R } 15

The dividend 543 sits in the allowed 540-549 window, so the digit is 3.

Answer: box = 3 (then 543 / 16 = 33 remainder 15)

4 · Reviewdoes it hold up?

Remainder 15 is less than the divisor 16, which is required. The dividend 543 is within 540-549, so the digit 3 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 3 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 540 through 549 by 16 and list the remainders; the largest, 15, occurs at 543, so box = 3.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 16 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 4 medium answer: box = 7 (then 587 / 21 = 27 remainder 20)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$58\square \div 21$

Show solution

1 · Understandwhat's really being asked

In the division 58box / 21, the box is a single digit 0-9 making 58box a number between 580 and 589. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 58box, where box is one digit from 0 to 9
The divisor is 21
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 20
The dividend ranges only from 580 to 589

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 21 can be at most 20, so we look for a dividend in the range 580-589 that leaves the biggest remainder. Finding the nearest multiple of 21 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 21 is 20; we hope to hit exactly remainder 20.

\text{remainder} < 21 \Rightarrow \text{max remainder} = 20

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 21 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 21 near 580. 21 x 27 = 567. Adding the leftover 20 gives 567 + 20 = 587, which is between 580 and 589.

21 \times 27 = 567,\quad 567 + 20 = 587

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 587 = 58box with box = 7, choose box = 7. Check: 587 / 21 = 27 remainder 20, the largest remainder available.

587 \div 21 = 27 \text{ R } 20

The dividend 587 sits in the allowed 580-589 window, so the digit is 7.

Answer: box = 7 (then 587 / 21 = 27 remainder 20)

4 · Reviewdoes it hold up?

Remainder 20 is less than the divisor 21, which is required. The dividend 587 is within 580-589, so the digit 7 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 7 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 580 through 589 by 21 and list the remainders; the largest, 20, occurs at 587, so box = 7.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 21 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 5 medium answer: box = 9 (then 639 / 19 = 33 remainder 12)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$63\square \div 19$

Show solution

1 · Understandwhat's really being asked

In the division 63box / 19, the box is a single digit 0-9 making 63box a number between 630 and 639. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 63box, where box is one digit from 0 to 9
The divisor is 19
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 18
The dividend ranges only from 630 to 639

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 19 can be at most 18, so we look for a dividend in the range 630-639 that leaves the biggest remainder. Finding the nearest multiple of 19 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 19 is 18; the digits available cannot reach 18, so we test each digit to find the largest remainder they do reach.

\text{remainder} < 19 \Rightarrow \text{max remainder} = 18

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 19 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 19 near 630. 19 x 33 = 627. Adding the leftover 12 gives 627 + 12 = 639, which is between 630 and 639.

19 \times 33 = 627,\quad 627 + 12 = 639

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 639 = 63box with box = 9, choose box = 9. Check: 639 / 19 = 33 remainder 12, the largest remainder available.

639 \div 19 = 33 \text{ R } 12

The dividend 639 sits in the allowed 630-639 window, so the digit is 9.

Answer: box = 9 (then 639 / 19 = 33 remainder 12)

4 · Reviewdoes it hold up?

Remainder 12 is less than the divisor 19, which is required. The dividend 639 is within 630-639, so the digit 9 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 9 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 630 through 639 by 19 and list the remainders; the largest, 12, occurs at 639, so box = 9.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 19 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 6 medium answer: box = 9 (then 699 / 31 = 22 remainder 17)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$69\square \div 31$

Show solution

1 · Understandwhat's really being asked

In the division 69box / 31, the box is a single digit 0-9 making 69box a number between 690 and 699. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 69box, where box is one digit from 0 to 9
The divisor is 31
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 30
The dividend ranges only from 690 to 699

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 31 can be at most 30, so we look for a dividend in the range 690-699 that leaves the biggest remainder. Finding the nearest multiple of 31 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 31 is 30; the digits available cannot reach 30, so we test each digit to find the largest remainder they do reach.

\text{remainder} < 31 \Rightarrow \text{max remainder} = 30

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 31 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 31 near 690. 31 x 22 = 682. Adding the leftover 17 gives 682 + 17 = 699, which is between 690 and 699.

31 \times 22 = 682,\quad 682 + 17 = 699

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 699 = 69box with box = 9, choose box = 9. Check: 699 / 31 = 22 remainder 17, the largest remainder available.

699 \div 31 = 22 \text{ R } 17

The dividend 699 sits in the allowed 690-699 window, so the digit is 9.

Answer: box = 9 (then 699 / 31 = 22 remainder 17)

4 · Reviewdoes it hold up?

Remainder 17 is less than the divisor 31, which is required. The dividend 699 is within 690-699, so the digit 9 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 9 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 690 through 699 by 31 and list the remainders; the largest, 17, occurs at 699, so box = 9.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 31 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 7 medium answer: box = 9 (then 769 / 18 = 42 remainder 13)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$76\square \div 18$

Show solution

1 · Understandwhat's really being asked

In the division 76box / 18, the box is a single digit 0-9 making 76box a number between 760 and 769. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 76box, where box is one digit from 0 to 9
The divisor is 18
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 17
The dividend ranges only from 760 to 769

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 18 can be at most 17, so we look for a dividend in the range 760-769 that leaves the biggest remainder. Finding the nearest multiple of 18 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 18 is 17; the digits available cannot reach 17, so we test each digit to find the largest remainder they do reach.

\text{remainder} < 18 \Rightarrow \text{max remainder} = 17

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 18 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 18 near 760. 18 x 42 = 756. Adding the leftover 13 gives 756 + 13 = 769, which is between 760 and 769.

18 \times 42 = 756,\quad 756 + 13 = 769

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 769 = 76box with box = 9, choose box = 9. Check: 769 / 18 = 42 remainder 13, the largest remainder available.

769 \div 18 = 42 \text{ R } 13

The dividend 769 sits in the allowed 760-769 window, so the digit is 9.

Answer: box = 9 (then 769 / 18 = 42 remainder 13)

4 · Reviewdoes it hold up?

Remainder 13 is less than the divisor 18, which is required. The dividend 769 is within 760-769, so the digit 9 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 9 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 760 through 769 by 18 and list the remainders; the largest, 13, occurs at 769, so box = 9.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 18 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 8 hard answer: box = 2 (then 782 / 27 = 28 remainder 26)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$78\square \div 27$

Show solution

1 · Understandwhat's really being asked

In the division 78box / 27, the box is a single digit 0-9 making 78box a number between 780 and 789. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 78box, where box is one digit from 0 to 9
The divisor is 27
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 26
The dividend ranges only from 780 to 789

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 27 can be at most 26, so we look for a dividend in the range 780-789 that leaves the biggest remainder. Finding the nearest multiple of 27 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 27 is 26; we hope to hit exactly remainder 26.

\text{remainder} < 27 \Rightarrow \text{max remainder} = 26

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 27 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 27 near 780. 27 x 28 = 756. Adding the leftover 26 gives 756 + 26 = 782, which is between 780 and 789.

27 \times 28 = 756,\quad 756 + 26 = 782

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 782 = 78box with box = 2, choose box = 2. Check: 782 / 27 = 28 remainder 26, the largest remainder available.

782 \div 27 = 28 \text{ R } 26

The dividend 782 sits in the allowed 780-789 window, so the digit is 2.

Answer: box = 2 (then 782 / 27 = 28 remainder 26)

4 · Reviewdoes it hold up?

Remainder 26 is less than the divisor 27, which is required. The dividend 782 is within 780-789, so the digit 2 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 2 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 780 through 789 by 27 and list the remainders; the largest, 26, occurs at 782, so box = 2.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 27 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 9 hard answer: box = 9 (then 829 / 29 = 28 remainder 17)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$82\square \div 29$

Show solution

1 · Understandwhat's really being asked

In the division 82box / 29, the box is a single digit 0-9 making 82box a number between 820 and 829. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 82box, where box is one digit from 0 to 9
The divisor is 29
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 28
The dividend ranges only from 820 to 829

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 29 can be at most 28, so we look for a dividend in the range 820-829 that leaves the biggest remainder. Finding the nearest multiple of 29 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 29 is 28; the digits available cannot reach 28, so we test each digit to find the largest remainder they do reach.

\text{remainder} < 29 \Rightarrow \text{max remainder} = 28

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 29 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 29 near 820. 29 x 28 = 812. Adding the leftover 17 gives 812 + 17 = 829, which is between 820 and 829.

29 \times 28 = 812,\quad 812 + 17 = 829

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 829 = 82box with box = 9, choose box = 9. Check: 829 / 29 = 28 remainder 17, the largest remainder available.

829 \div 29 = 28 \text{ R } 17

The dividend 829 sits in the allowed 820-829 window, so the digit is 9.

Answer: box = 9 (then 829 / 29 = 28 remainder 17)

4 · Reviewdoes it hold up?

Remainder 17 is less than the divisor 29, which is required. The dividend 829 is within 820-829, so the digit 9 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 9 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 820 through 829 by 29 and list the remainders; the largest, 17, occurs at 829, so box = 9.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 29 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!

Variant 10 hard answer: box = 9 (then 919 / 23 = 39 remainder 22)

The box $\square$ can be filled with any digit from $0$ to $9$ . In the division below, find the digit for $\square$ that makes the remainder as large as possible.

$91\square \div 23$

Show solution

1 · Understandwhat's really being asked

In the division 91box / 23, the box is a single digit 0-9 making 91box a number between 910 and 919. Choose the digit that makes the remainder as large as it can be.

Givens

The dividend is 91box, where box is one digit from 0 to 9
The divisor is 23
We want the remainder to be as large as possible

Unknowns

The digit box that maximizes the remainder

Constraints

A remainder must be less than the divisor, so the remainder can be at most 22
The dividend ranges only from 910 to 919

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

The remainder when dividing by 23 can be at most 22, so we look for a dividend in the range 910-919 that leaves the biggest remainder. Finding the nearest multiple of 23 below this range and stepping up turns it into a simple check of a single digit.

3 · Execute3 carry out the plan

1Find the largest possible remainder

#9 Solve an Easier Related Problem 4.NBT.B.6

Because the remainder is always smaller than the divisor, the biggest remainder possible when dividing by 23 is 22; we hope to hit exactly remainder 22.

\text{remainder} < 23 \Rightarrow \text{max remainder} = 22

Knowing the remainder caps at one less than the divisor turns this into a target hunt.

2Locate the multiple of 23 inside the range

#6 Guess and Check 4.NBT.B.6

Find a multiple of 23 near 910. 23 x 39 = 897. Adding the leftover 22 gives 897 + 22 = 919, which is between 910 and 919.

23 \times 39 = 897,\quad 897 + 22 = 919

Adding the biggest reachable remainder to a known multiple shows exactly which dividend works.

3Read off the digit

#6 Guess and Check 4.NBT.B.6

Since 919 = 91box with box = 9, choose box = 9. Check: 919 / 23 = 39 remainder 22, the largest remainder available.

919 \div 23 = 39 \text{ R } 22

The dividend 919 sits in the allowed 910-919 window, so the digit is 9.

Answer: box = 9 (then 919 / 23 = 39 remainder 22)

4 · Reviewdoes it hold up?

Remainder 22 is less than the divisor 23, which is required. The dividend 919 is within 910-919, so the digit 9 is valid. Checking the other digits 0-9 gives no larger remainder, confirming 9 is the right choice.

Another way: Guess and Check (tool 6) every digit: divide 910 through 919 by 23 and list the remainders; the largest, 22, occurs at 919, so box = 9.

Standardsmin grade 4

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends — Dividing by 23 to compute quotients and remainders and using the remainder-less-than-divisor rule.

💡Takeaway. This only needs the Grade 4 fact that a remainder is always smaller than the divisor -- aim for the biggest remainder and find the matching digit!