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← 3-1 · Recover missing digits from column carries · Recover Hidden Digits from Carries

Recover missing digits from column carries · 10 practice problems

3.NBT.A.33.OA.B.5

Generated variants — 10

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

Variant 1 answer: A = 3, B = 7

In the multiplication below, find the digits AA and BB.

A B×4148\begin{array}{r} A\ B \\ \times\quad 4 \\ \hline 1 4 8 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 44 equals 148148.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 4, and the product is 148. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 4 equals 148
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 148 by 4 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 4 equals 148, divide 148 by 4 to recover the number.
148÷4=37148 \div 4 = 37
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 37, so the tens digit A is 3 and the ones digit B is 7.
37A=3, B=737 \Rightarrow A = 3,\ B = 7
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 4 times 7 = 28, write 8 carry 2; 4 times 3 = 12, plus the carry 2 = 14. That builds 148.
37×4=14837 \times 4 = 148
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 3, B = 7

Review

37 times 4 = 148 exactly, and both A=3 and B=7 are valid single digits with A not zero, so the two-digit number 37 is legitimate.

Guess and check the ones column first (tool 6): 4 times B must end in 8, which narrows B; only B = 7 with the carry produces 148, then the tens column forces A = 3.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 148 by 4 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 2 answer: A = 2, B = 9

In the multiplication below, find the digits AA and BB.

A B×5145\begin{array}{r} A\ B \\ \times\quad 5 \\ \hline 1 4 5 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 55 equals 145145.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 5, and the product is 145. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 5 equals 145
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 145 by 5 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 5 equals 145, divide 145 by 5 to recover the number.
145÷5=29145 \div 5 = 29
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 29, so the tens digit A is 2 and the ones digit B is 9.
29A=2, B=929 \Rightarrow A = 2,\ B = 9
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 5 times 9 = 45, write 5 carry 4; 5 times 2 = 10, plus the carry 4 = 14. That builds 145.
29×5=14529 \times 5 = 145
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 2, B = 9

Review

29 times 5 = 145 exactly, and both A=2 and B=9 are valid single digits with A not zero, so the two-digit number 29 is legitimate.

Guess and check the ones column first (tool 6): 5 times B must end in 5, which narrows B; only B = 9 with the carry produces 145, then the tens column forces A = 2.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 145 by 5 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 3 answer: A = 4, B = 6

In the multiplication below, find the digits AA and BB.

A B×7322\begin{array}{r} A\ B \\ \times\quad 7 \\ \hline 3 2 2 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 77 equals 322322.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 7, and the product is 322. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 7 equals 322
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 322 by 7 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 7 equals 322, divide 322 by 7 to recover the number.
322÷7=46322 \div 7 = 46
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 46, so the tens digit A is 4 and the ones digit B is 6.
46A=4, B=646 \Rightarrow A = 4,\ B = 6
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 7 times 6 = 42, write 2 carry 4; 7 times 4 = 28, plus the carry 4 = 32. That builds 322.
46×7=32246 \times 7 = 322
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 4, B = 6

Review

46 times 7 = 322 exactly, and both A=4 and B=6 are valid single digits with A not zero, so the two-digit number 46 is legitimate.

Guess and check the ones column first (tool 6): 7 times B must end in 2, which narrows B; only B = 6 with the carry produces 322, then the tens column forces A = 4.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 322 by 7 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 4 answer: A = 2, B = 7

In the multiplication below, find the digits AA and BB.

A B×8216\begin{array}{r} A\ B \\ \times\quad 8 \\ \hline 2 1 6 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 88 equals 216216.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 8, and the product is 216. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 8 equals 216
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 216 by 8 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 8 equals 216, divide 216 by 8 to recover the number.
216÷8=27216 \div 8 = 27
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 27, so the tens digit A is 2 and the ones digit B is 7.
27A=2, B=727 \Rightarrow A = 2,\ B = 7
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 8 times 7 = 56, write 6 carry 5; 8 times 2 = 16, plus the carry 5 = 21. That builds 216.
27×8=21627 \times 8 = 216
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 2, B = 7

Review

27 times 8 = 216 exactly, and both A=2 and B=7 are valid single digits with A not zero, so the two-digit number 27 is legitimate.

Guess and check the ones column first (tool 6): 8 times B must end in 6, which narrows B; only B = 7 with the carry produces 216, then the tens column forces A = 2.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 216 by 8 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 5 answer: A = 7, B = 4

In the multiplication below, find the digits AA and BB.

A B×6444\begin{array}{r} A\ B \\ \times\quad 6 \\ \hline 4 4 4 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 66 equals 444444.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 6, and the product is 444. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 6 equals 444
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 444 by 6 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 6 equals 444, divide 444 by 6 to recover the number.
444÷6=74444 \div 6 = 74
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 74, so the tens digit A is 7 and the ones digit B is 4.
74A=7, B=474 \Rightarrow A = 7,\ B = 4
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 6 times 4 = 24, write 4 carry 2; 6 times 7 = 42, plus the carry 2 = 44. That builds 444.
74×6=44474 \times 6 = 444
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 7, B = 4

Review

74 times 6 = 444 exactly, and both A=7 and B=4 are valid single digits with A not zero, so the two-digit number 74 is legitimate.

Guess and check the ones column first (tool 6): 6 times B must end in 4, which narrows B; only B = 4 with the carry produces 444, then the tens column forces A = 7.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 444 by 6 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 6 answer: A = 5, B = 2

In the multiplication below, find the digits AA and BB.

A B×3156\begin{array}{r} A\ B \\ \times\quad 3 \\ \hline 1 5 6 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 33 equals 156156.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 3, and the product is 156. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 3 equals 156
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 156 by 3 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 3 equals 156, divide 156 by 3 to recover the number.
156÷3=52156 \div 3 = 52
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 52, so the tens digit A is 5 and the ones digit B is 2.
52A=5, B=252 \Rightarrow A = 5,\ B = 2
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 3 times 2 = 6 (no carry); 3 times 5 = 15. That builds 156.
52×3=15652 \times 3 = 156
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 5, B = 2

Review

52 times 3 = 156 exactly, and both A=5 and B=2 are valid single digits with A not zero, so the two-digit number 52 is legitimate.

Guess and check the ones column first (tool 6): 3 times B must end in 6, which narrows B; only B = 2 with the carry produces 156, then the tens column forces A = 5.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 156 by 3 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 7 answer: A = 8, B = 5

In the multiplication below, find the digits AA and BB.

A B×3255\begin{array}{r} A\ B \\ \times\quad 3 \\ \hline 2 5 5 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 33 equals 255255.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 3, and the product is 255. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 3 equals 255
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 255 by 3 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 3 equals 255, divide 255 by 3 to recover the number.
255÷3=85255 \div 3 = 85
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 85, so the tens digit A is 8 and the ones digit B is 5.
85A=8, B=585 \Rightarrow A = 8,\ B = 5
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 3 times 5 = 15, write 5 carry 1; 3 times 8 = 24, plus the carry 1 = 25. That builds 255.
85×3=25585 \times 3 = 255
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 8, B = 5

Review

85 times 3 = 255 exactly, and both A=8 and B=5 are valid single digits with A not zero, so the two-digit number 85 is legitimate.

Guess and check the ones column first (tool 6): 3 times B must end in 5, which narrows B; only B = 5 with the carry produces 255, then the tens column forces A = 8.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 255 by 3 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 8 answer: A = 6, B = 3

In the multiplication below, find the digits AA and BB.

A B×2126\begin{array}{r} A\ B \\ \times\quad 2 \\ \hline 1 2 6 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 22 equals 126126.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 2, and the product is 126. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 2 equals 126
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 126 by 2 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 2 equals 126, divide 126 by 2 to recover the number.
126÷2=63126 \div 2 = 63
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 63, so the tens digit A is 6 and the ones digit B is 3.
63A=6, B=363 \Rightarrow A = 6,\ B = 3
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 2 times 3 = 6 (no carry); 2 times 6 = 12. That builds 126.
63×2=12663 \times 2 = 126
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 6, B = 3

Review

63 times 2 = 126 exactly, and both A=6 and B=3 are valid single digits with A not zero, so the two-digit number 63 is legitimate.

Guess and check the ones column first (tool 6): 2 times B must end in 6, which narrows B; only B = 3 with the carry produces 126, then the tens column forces A = 6.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 126 by 2 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 9 answer: A = 4, B = 8

In the multiplication below, find the digits AA and BB.

A B×6288\begin{array}{r} A\ B \\ \times\quad 6 \\ \hline 2 8 8 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 66 equals 288288.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 6, and the product is 288. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 6 equals 288
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 288 by 6 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 6 equals 288, divide 288 by 6 to recover the number.
288÷6=48288 \div 6 = 48
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 48, so the tens digit A is 4 and the ones digit B is 8.
48A=4, B=848 \Rightarrow A = 4,\ B = 8
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 6 times 8 = 48, write 8 carry 4; 6 times 4 = 24, plus the carry 4 = 28. That builds 288.
48×6=28848 \times 6 = 288
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 4, B = 8

Review

48 times 6 = 288 exactly, and both A=4 and B=8 are valid single digits with A not zero, so the two-digit number 48 is legitimate.

Guess and check the ones column first (tool 6): 6 times B must end in 8, which narrows B; only B = 8 with the carry produces 288, then the tens column forces A = 4.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 288 by 6 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!
Variant 10 answer: A = 5, B = 8

In the multiplication below, find the digits AA and BB.

A B×4232\begin{array}{r} A\ B \\ \times\quad 4 \\ \hline 2 3 2 \end{array}

Here AA is the tens digit and BB is the ones digit of the two-digit number being multiplied. The two-digit number times 44 equals 232232.

Show solution

Understand

A two-digit number with tens digit A and ones digit B is multiplied by 4, and the product is 232. We must find the digits A and B.

Givens
  • The two-digit number is written A B (A tens, B ones)
  • The two-digit number times 4 equals 232
Unknowns
  • The tens digit A
  • The ones digit B
Constraints
  • A and B are single digits (0-9), and A is not 0 since A B is a two-digit number

Plan

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

The result of the multiplication is given, so we undo it: dividing 232 by 4 recovers the two-digit number, and reading its digits gives A and B. A quick column check confirms the carry.

Execute

#11 Work Backwards 3.OA.A.3
Since the two-digit number times 4 equals 232, divide 232 by 4 to recover the number.
232÷4=58232 \div 4 = 58
Multiplication and division are opposites, so dividing the product by the multiplier gives back the original number.
#11 Work Backwards 3.NBT.A.3
The recovered number is 58, so the tens digit A is 5 and the ones digit B is 8.
58A=5, B=858 \Rightarrow A = 5,\ B = 8
Place value tells us the left digit is the tens and the right digit is the ones.
#6 Guess and Check 3.OA.B.5
Verify by multiplying: 4 times 8 = 32, write 2 carry 3; 4 times 5 = 20, plus the carry 3 = 23. That builds 232.
58×4=23258 \times 4 = 232
Rebuilding the product column by column confirms the carries line up exactly.
Answer: A = 5, B = 8

Review

58 times 4 = 232 exactly, and both A=5 and B=8 are valid single digits with A not zero, so the two-digit number 58 is legitimate.

Guess and check the ones column first (tool 6): 4 times B must end in 2, which narrows B; only B = 8 with the carry produces 232, then the tens column forces A = 5.

Standards · min grade 3

  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Dividing 232 by 4 to recover the two-digit factor
  • 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Using place value to read the tens digit A and ones digit B
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Checking the column multiplication with carrying
💡 When you know the answer to a times problem, just divide to walk backwards to the hidden number, then read its digits!