← 3-2 · Fill long-division blanks from the known · Recover Hidden Digits from Carries

Fill long-division blanks from the known · 7 practice problems

3.OA.C.73.OA.B.6

Generated variants — 7

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

Variant 1 answer: Dividend 69 and quotient 17: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 4, bring-down row = 29, final subtraction row = 28 (remainder 1).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 4, and the dividend is a two-digit number whose tens digit is 6 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 7. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 1. Find every hidden digit.

Show solution

Understand

In a long-division layout, 4 divides a two-digit dividend 6-blank, giving a two-digit quotient blank-7 with remainder 1. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 4.
The dividend is 6 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 7 in the ones place.
The final remainder is 1.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 1 is less than the divisor 4.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 4 times quotient plus 1 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 4 times the quotient plus the remainder 1. The quotient ends in 7 and the dividend is in the 60s, so the quotient is 17, giving dividend 4 times 17 plus 1.

4 \times 17 + 1 = 68 + 1 = 69

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 69, so its hidden ones digit is 9. The quotient is 17, so its hidden tens digit is 1.

69 \div 4 = 17 \cdots 1

Once the dividend is 69, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 4 goes into 6 1 time(s). The quotient tens digit 1 times divisor 4 is 4, the one-box partial product. Subtracting 4 from 6 leaves 2.

1 \times 4 = 4,\quad 6 - 4 = 2

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 9 to make 29, the two-box bring-down row. The ones quotient digit 7 times 4 is 28, the two-box subtraction row. Subtracting gives the remainder 1.

29 - 28 = 1,\quad 7 \times 4 = 28

The bring-down row 29 and the product 28 differ by exactly the remainder 1, confirming all boxes.

Answer: Dividend 69 and quotient 17: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 4, bring-down row = 29, final subtraction row = 28 (remainder 1).

Review

Check by dividing: 69 divided by 4 is 17 remainder 1, since 4 times 17 is 68 and 69 minus 68 is 1, and 1 is less than 4. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 4 times Q plus 1 = 6_ where Q ends in 7; only Q = 17 keeps the dividend two digits starting with 6.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 69 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 2 answer: Dividend 92 and quotient 18: dividend ones digit = 2, quotient tens digit = 1, tens partial product = 5, bring-down row = 42, final subtraction row = 40 (remainder 2).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 5, and the dividend is a two-digit number whose tens digit is 9 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 8. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 2. Find every hidden digit.

Show solution

Understand

In a long-division layout, 5 divides a two-digit dividend 9-blank, giving a two-digit quotient blank-8 with remainder 2. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 5.
The dividend is 9 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 8 in the ones place.
The final remainder is 2.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 2 is less than the divisor 5.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 5 times quotient plus 2 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 5 times the quotient plus the remainder 2. The quotient ends in 8 and the dividend is in the 90s, so the quotient is 18, giving dividend 5 times 18 plus 2.

5 \times 18 + 2 = 90 + 2 = 92

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 92, so its hidden ones digit is 2. The quotient is 18, so its hidden tens digit is 1.

92 \div 5 = 18 \cdots 2

Once the dividend is 92, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 5 goes into 9 1 time(s). The quotient tens digit 1 times divisor 5 is 5, the one-box partial product. Subtracting 5 from 9 leaves 4.

1 \times 5 = 5,\quad 9 - 5 = 4

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 2 to make 42, the two-box bring-down row. The ones quotient digit 8 times 5 is 40, the two-box subtraction row. Subtracting gives the remainder 2.

42 - 40 = 2,\quad 8 \times 5 = 40

The bring-down row 42 and the product 40 differ by exactly the remainder 2, confirming all boxes.

Answer: Dividend 92 and quotient 18: dividend ones digit = 2, quotient tens digit = 1, tens partial product = 5, bring-down row = 42, final subtraction row = 40 (remainder 2).

Review

Check by dividing: 92 divided by 5 is 18 remainder 2, since 5 times 18 is 90 and 92 minus 90 is 2, and 2 is less than 5. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 5 times Q plus 2 = 9_ where Q ends in 8; only Q = 18 keeps the dividend two digits starting with 9.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 92 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 3 answer: Dividend 69 and quotient 13: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 5, bring-down row = 19, final subtraction row = 15 (remainder 4).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 5, and the dividend is a two-digit number whose tens digit is 6 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 3. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 4. Find every hidden digit.

Show solution

Understand

In a long-division layout, 5 divides a two-digit dividend 6-blank, giving a two-digit quotient blank-3 with remainder 4. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 5.
The dividend is 6 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 3 in the ones place.
The final remainder is 4.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 4 is less than the divisor 5.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 5 times quotient plus 4 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 5 times the quotient plus the remainder 4. The quotient ends in 3 and the dividend is in the 60s, so the quotient is 13, giving dividend 5 times 13 plus 4.

5 \times 13 + 4 = 65 + 4 = 69

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 69, so its hidden ones digit is 9. The quotient is 13, so its hidden tens digit is 1.

69 \div 5 = 13 \cdots 4

Once the dividend is 69, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 5 goes into 6 1 time(s). The quotient tens digit 1 times divisor 5 is 5, the one-box partial product. Subtracting 5 from 6 leaves 1.

1 \times 5 = 5,\quad 6 - 5 = 1

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 9 to make 19, the two-box bring-down row. The ones quotient digit 3 times 5 is 15, the two-box subtraction row. Subtracting gives the remainder 4.

19 - 15 = 4,\quad 3 \times 5 = 15

The bring-down row 19 and the product 15 differ by exactly the remainder 4, confirming all boxes.

Answer: Dividend 69 and quotient 13: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 5, bring-down row = 19, final subtraction row = 15 (remainder 4).

Review

Check by dividing: 69 divided by 5 is 13 remainder 4, since 5 times 13 is 65 and 69 minus 65 is 4, and 4 is less than 5. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 5 times Q plus 4 = 6_ where Q ends in 3; only Q = 13 keeps the dividend two digits starting with 6.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 69 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 4 answer: Dividend 67 and quotient 16: dividend ones digit = 7, quotient tens digit = 1, tens partial product = 4, bring-down row = 27, final subtraction row = 24 (remainder 3).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 4, and the dividend is a two-digit number whose tens digit is 6 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 6. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 3. Find every hidden digit.

Show solution

Understand

In a long-division layout, 4 divides a two-digit dividend 6-blank, giving a two-digit quotient blank-6 with remainder 3. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 4.
The dividend is 6 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 6 in the ones place.
The final remainder is 3.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 3 is less than the divisor 4.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 4 times quotient plus 3 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 4 times the quotient plus the remainder 3. The quotient ends in 6 and the dividend is in the 60s, so the quotient is 16, giving dividend 4 times 16 plus 3.

4 \times 16 + 3 = 64 + 3 = 67

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 67, so its hidden ones digit is 7. The quotient is 16, so its hidden tens digit is 1.

67 \div 4 = 16 \cdots 3

Once the dividend is 67, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 4 goes into 6 1 time(s). The quotient tens digit 1 times divisor 4 is 4, the one-box partial product. Subtracting 4 from 6 leaves 2.

1 \times 4 = 4,\quad 6 - 4 = 2

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 7 to make 27, the two-box bring-down row. The ones quotient digit 6 times 4 is 24, the two-box subtraction row. Subtracting gives the remainder 3.

27 - 24 = 3,\quad 6 \times 4 = 24

The bring-down row 27 and the product 24 differ by exactly the remainder 3, confirming all boxes.

Answer: Dividend 67 and quotient 16: dividend ones digit = 7, quotient tens digit = 1, tens partial product = 4, bring-down row = 27, final subtraction row = 24 (remainder 3).

Review

Check by dividing: 67 divided by 4 is 16 remainder 3, since 4 times 16 is 64 and 67 minus 64 is 3, and 3 is less than 4. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 4 times Q plus 3 = 6_ where Q ends in 6; only Q = 16 keeps the dividend two digits starting with 6.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 67 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 5 answer: Dividend 89 and quotient 14: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 6, bring-down row = 29, final subtraction row = 24 (remainder 5).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 6, and the dividend is a two-digit number whose tens digit is 8 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 4. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 5. Find every hidden digit.

Show solution

Understand

In a long-division layout, 6 divides a two-digit dividend 8-blank, giving a two-digit quotient blank-4 with remainder 5. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 6.
The dividend is 8 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 4 in the ones place.
The final remainder is 5.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 5 is less than the divisor 6.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 6 times quotient plus 5 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 6 times the quotient plus the remainder 5. The quotient ends in 4 and the dividend is in the 80s, so the quotient is 14, giving dividend 6 times 14 plus 5.

6 \times 14 + 5 = 84 + 5 = 89

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 89, so its hidden ones digit is 9. The quotient is 14, so its hidden tens digit is 1.

89 \div 6 = 14 \cdots 5

Once the dividend is 89, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 6 goes into 8 1 time(s). The quotient tens digit 1 times divisor 6 is 6, the one-box partial product. Subtracting 6 from 8 leaves 2.

1 \times 6 = 6,\quad 8 - 6 = 2

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 9 to make 29, the two-box bring-down row. The ones quotient digit 4 times 6 is 24, the two-box subtraction row. Subtracting gives the remainder 5.

29 - 24 = 5,\quad 4 \times 6 = 24

The bring-down row 29 and the product 24 differ by exactly the remainder 5, confirming all boxes.

Answer: Dividend 89 and quotient 14: dividend ones digit = 9, quotient tens digit = 1, tens partial product = 6, bring-down row = 29, final subtraction row = 24 (remainder 5).

Review

Check by dividing: 89 divided by 6 is 14 remainder 5, since 6 times 14 is 84 and 89 minus 84 is 5, and 5 is less than 6. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 6 times Q plus 5 = 8_ where Q ends in 4; only Q = 14 keeps the dividend two digits starting with 8.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 89 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 6 answer: Dividend 56 and quotient 18: dividend ones digit = 6, quotient tens digit = 1, tens partial product = 3, bring-down row = 26, final subtraction row = 24 (remainder 2).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 3, and the dividend is a two-digit number whose tens digit is 5 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 8. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 2. Find every hidden digit.

Show solution

Understand

In a long-division layout, 3 divides a two-digit dividend 5-blank, giving a two-digit quotient blank-8 with remainder 2. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 3.
The dividend is 5 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 8 in the ones place.
The final remainder is 2.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 2 is less than the divisor 3.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 3 times quotient plus 2 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 3 times the quotient plus the remainder 2. The quotient ends in 8 and the dividend is in the 50s, so the quotient is 18, giving dividend 3 times 18 plus 2.

3 \times 18 + 2 = 54 + 2 = 56

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 56, so its hidden ones digit is 6. The quotient is 18, so its hidden tens digit is 1.

56 \div 3 = 18 \cdots 2

Once the dividend is 56, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 3 goes into 5 1 time(s). The quotient tens digit 1 times divisor 3 is 3, the one-box partial product. Subtracting 3 from 5 leaves 2.

1 \times 3 = 3,\quad 5 - 3 = 2

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 6 to make 26, the two-box bring-down row. The ones quotient digit 8 times 3 is 24, the two-box subtraction row. Subtracting gives the remainder 2.

26 - 24 = 2,\quad 8 \times 3 = 24

The bring-down row 26 and the product 24 differ by exactly the remainder 2, confirming all boxes.

Answer: Dividend 56 and quotient 18: dividend ones digit = 6, quotient tens digit = 1, tens partial product = 3, bring-down row = 26, final subtraction row = 24 (remainder 2).

Review

Check by dividing: 56 divided by 3 is 18 remainder 2, since 3 times 18 is 54 and 56 minus 54 is 2, and 2 is less than 3. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 3 times Q plus 2 = 5_ where Q ends in 8; only Q = 18 keeps the dividend two digits starting with 5.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 56 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!

Variant 7 answer: Dividend 90 and quotient 12: dividend ones digit = 0, quotient tens digit = 1, tens partial product = 7, bring-down row = 20, final subtraction row = 14 (remainder 6).

Fill in each $\square$ with the correct digit.

The figure below is a long division. The divisor is 7, and the dividend is a two-digit number whose tens digit is 9 and whose ones digit is hidden by a $\square$ . The quotient is a two-digit number whose tens digit is hidden by a $\square$ and whose ones digit is 2. In the worked layout, the partial product of the tens digit (a one-digit number), the two-digit number brought down after subtracting, and the final two-digit number being subtracted each have their digits hidden by $\square$ , and the final remainder is 6. Find every hidden digit.

Show solution

Understand

In a long-division layout, 7 divides a two-digit dividend 9-blank, giving a two-digit quotient blank-2 with remainder 6. We must find every hidden digit, including the partial-product and subtraction rows.

Givens

The divisor is 7.
The dividend is 9 in the tens place and a hidden digit in the ones place.
The quotient is a hidden tens digit and 2 in the ones place.
The final remainder is 6.
The figure shows a one-box partial product, a two-box bring-down row, and a two-box subtraction row.

Unknowns

The dividend's ones digit, the quotient's tens digit, and every hidden box in the worked layout.

Constraints

Dividend = divisor times quotient plus remainder.
Every box is a single digit; the remainder 6 is less than the divisor 7.

Plan

#11 Work Backwards · also uses: #6 Guess and Check

Use the rule dividend = 7 times quotient plus 6 to recover the dividend, then replay the long division step by step to fill each hidden box.

Execute

#11 Work Backwards 3.OA.B.6

The dividend equals 7 times the quotient plus the remainder 6. The quotient ends in 2 and the dividend is in the 90s, so the quotient is 12, giving dividend 7 times 12 plus 6.

7 \times 12 + 6 = 84 + 6 = 90

Multiplying back from quotient and remainder is the inverse of dividing, so it pins down the dividend exactly.

#6 Guess and Check 3.OA.C.7

The dividend is 90, so its hidden ones digit is 0. The quotient is 12, so its hidden tens digit is 1.

90 \div 7 = 12 \cdots 6

Once the dividend is 90, reading its ones digit and the quotient's tens digit is direct.

#11 Work Backwards 3.OA.C.7

First 7 goes into 9 1 time(s). The quotient tens digit 1 times divisor 7 is 7, the one-box partial product. Subtracting 7 from 9 leaves 2.

1 \times 7 = 7,\quad 9 - 7 = 2

Each long-division step multiplies the new quotient digit by the divisor, so the partial product is fixed.

#6 Guess and Check 3.OA.C.7

Bring down the 0 to make 20, the two-box bring-down row. The ones quotient digit 2 times 7 is 14, the two-box subtraction row. Subtracting gives the remainder 6.

20 - 14 = 6,\quad 2 \times 7 = 14

The bring-down row 20 and the product 14 differ by exactly the remainder 6, confirming all boxes.

Answer: Dividend 90 and quotient 12: dividend ones digit = 0, quotient tens digit = 1, tens partial product = 7, bring-down row = 20, final subtraction row = 14 (remainder 6).

Review

Check by dividing: 90 divided by 7 is 12 remainder 6, since 7 times 12 is 84 and 90 minus 84 is 6, and 6 is less than 7. All digits are consistent.

Convert to a missing-factor equation (tool 13): solve 7 times Q plus 6 = 9_ where Q ends in 2; only Q = 12 keeps the dividend two digits starting with 9.

Standards · min grade 3

3.OA.B.6 Understand division as an unknown-factor problem — Recovering the dividend 90 from the quotient and remainder by multiplying back.
3.OA.C.7 Fluently multiply and divide within 100 — Replaying each long-division step to fill the partial-product and subtraction boxes.

💡 This only needs Grade 3 division facts: multiply the quotient back to get the dividend, then redo the steps to fill every box!