Practice · The remainder is always less than the divisor · Sensim Math

Variant 1 easy

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.

$37\square \div 14$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 37

\square

with

\square

= 7, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$45\square \div 13$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 45

\square

with

\square

= 4, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$54\square \div 16$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 54

\square

with

\square

= 3, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$58\square \div 21$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 58

\square

with

\square

= 7, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$63\square \div 19$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 63

\square

with

\square

= 9, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$69\square \div 31$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 69

\square

with

\square

= 9, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$76\square \div 18$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 76

\square

with

\square

= 9, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$78\square \div 27$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 78

\square

with

\square

= 2, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$82\square \div 29$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 82

\square

with

\square

= 9, choose

\square

= 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:

\square

= 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 $\square$ = 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

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.

$91\square \div 23$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

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

Unknowns

The digit $\square$ 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 = 91

\square

with

\square

= 9, choose

\square

= 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:

\square

= 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 $\square$ = 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!

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

Generated variants — 10

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?