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Deduce hidden digits from the ones place up · 11 practice problems

4.NBT.B.5

Generated variants — 11

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

Variant 1 answer: A = 7, B = 8, C = 0, D = 5, E = 1

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

3 A 5×4 B3 C 0 01 D 0 00E 8 0 0 0\begin{array}{r} 3\,A\,5 \\ \times \quad 4\,B \\ \hline 3\,C\,0\,0 \\ 1\,D\,0\,0 \phantom{0} \\ \hline E\,8\,0\,0\,0 \end{array}

(Layout) The top three-digit number is 3 A 53\,A\,5 and the two-digit multiplier is 4 B4\,B. The first partial product is 3 C 0 03\,C\,0\,0, the second partial product is 1 D 0 01\,D\,0\,0 written shifted one place to the left, and adding the two partial products gives the final product E 8 0 0 0E\,8\,0\,0\,0.

Show solution

Understand

A three-digit number 3A5 is multiplied by a two-digit number 4B. The two partial products are shown (the second shifted one place left), and the final product is E8000. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 3A5 (a three-digit number)
  • Bottom factor is 4B (a two-digit number)
  • First partial product (top factor times ones digit B) is 3C00
  • Second partial product (top factor times tens digit 4) is 1D00, shifted one place left
  • Final product is E8000
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 4
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 4, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 4. So the top number times 4 must equal 1500. Testing digits, 375 x 4 = 1500, which matches the pattern with A = 7 and D = 5.
375×4=1500⇒A=7, D=5375 \times 4 = 1500 \Rightarrow A=7,\ D=5
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 375 multiplied by the ones digit B and equals 3000. Since 375 x 8 = 3000, we get B = 8 and C = 0.
375×8=3000⇒B=8, C=0375 \times 8 = 3000 \Rightarrow B=8,\ C=0
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 48. Compute 375 x 48 = 3000 + 15000 = 18000, which matches the product with E = 1.
375×48=18000⇒E=1375 \times 48 = 18000 \Rightarrow E=1
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 7, B = 8, C = 0, D = 5, E = 1

Review

Check the whole multiplication: 375 x 48 = 18000, a five-digit number matching the product. Both partial products (3000 and 1500 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (375 x 4 and 375 x 8) and the full product 375 x 48 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 2 answer: A = 1, B = 4, C = 2, D = 6, E = 1

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

8 A 7×2 B3 C 6 81 D 3 40E 9 6 0 8\begin{array}{r} 8\,A\,7 \\ \times \quad 2\,B \\ \hline 3\,C\,6\,8 \\ 1\,D\,3\,4 \phantom{0} \\ \hline E\,9\,6\,0\,8 \end{array}

(Layout) The top three-digit number is 8 A 78\,A\,7 and the two-digit multiplier is 2 B2\,B. The first partial product is 3 C 6 83\,C\,6\,8, the second partial product is 1 D 3 41\,D\,3\,4 written shifted one place to the left, and adding the two partial products gives the final product E 9 6 0 8E\,9\,6\,0\,8.

Show solution

Understand

A three-digit number 8A7 is multiplied by a two-digit number 2B. The two partial products are shown (the second shifted one place left), and the final product is E9608. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 8A7 (a three-digit number)
  • Bottom factor is 2B (a two-digit number)
  • First partial product (top factor times ones digit B) is 3C68
  • Second partial product (top factor times tens digit 2) is 1D34, shifted one place left
  • Final product is E9608
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 2
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 2, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 2. So the top number times 2 must equal 1634. Testing digits, 817 x 2 = 1634, which matches the pattern with A = 1 and D = 6.
817×2=1634⇒A=1, D=6817 \times 2 = 1634 \Rightarrow A=1,\ D=6
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 817 multiplied by the ones digit B and equals 3268. Since 817 x 4 = 3268, we get B = 4 and C = 2.
817×4=3268⇒B=4, C=2817 \times 4 = 3268 \Rightarrow B=4,\ C=2
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 24. Compute 817 x 24 = 3268 + 16340 = 19608, which matches the product with E = 1.
817×24=19608⇒E=1817 \times 24 = 19608 \Rightarrow E=1
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 1, B = 4, C = 2, D = 6, E = 1

Review

Check the whole multiplication: 817 x 24 = 19608, a five-digit number matching the product. Both partial products (3268 and 1634 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (817 x 2 and 817 x 4) and the full product 817 x 24 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 3 answer: A = 2, B = 4, C = 5, D = 8, E = 2

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

6 A 8×3 B2 C 1 21 D 8 40E 1 3 5 2\begin{array}{r} 6\,A\,8 \\ \times \quad 3\,B \\ \hline 2\,C\,1\,2 \\ 1\,D\,8\,4 \phantom{0} \\ \hline E\,1\,3\,5\,2 \end{array}

(Layout) The top three-digit number is 6 A 86\,A\,8 and the two-digit multiplier is 3 B3\,B. The first partial product is 2 C 1 22\,C\,1\,2, the second partial product is 1 D 8 41\,D\,8\,4 written shifted one place to the left, and adding the two partial products gives the final product E 1 3 5 2E\,1\,3\,5\,2.

Show solution

Understand

A three-digit number 6A8 is multiplied by a two-digit number 3B. The two partial products are shown (the second shifted one place left), and the final product is E1352. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 6A8 (a three-digit number)
  • Bottom factor is 3B (a two-digit number)
  • First partial product (top factor times ones digit B) is 2C12
  • Second partial product (top factor times tens digit 3) is 1D84, shifted one place left
  • Final product is E1352
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 3
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 3, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 3. So the top number times 3 must equal 1884. Testing digits, 628 x 3 = 1884, which matches the pattern with A = 2 and D = 8.
628×3=1884⇒A=2, D=8628 \times 3 = 1884 \Rightarrow A=2,\ D=8
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 628 multiplied by the ones digit B and equals 2512. Since 628 x 4 = 2512, we get B = 4 and C = 5.
628×4=2512⇒B=4, C=5628 \times 4 = 2512 \Rightarrow B=4,\ C=5
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 34. Compute 628 x 34 = 2512 + 18840 = 21352, which matches the product with E = 2.
628×34=21352⇒E=2628 \times 34 = 21352 \Rightarrow E=2
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 2, B = 4, C = 5, D = 8, E = 2

Review

Check the whole multiplication: 628 x 34 = 21352, a five-digit number matching the product. Both partial products (2512 and 1884 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (628 x 3 and 628 x 4) and the full product 628 x 34 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 4 answer: A = 9, B = 7, C = 1, D = 7, E = 2

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

5 A 6×3 B4 C 7 21 D 8 80E 2 0 5 2\begin{array}{r} 5\,A\,6 \\ \times \quad 3\,B \\ \hline 4\,C\,7\,2 \\ 1\,D\,8\,8 \phantom{0} \\ \hline E\,2\,0\,5\,2 \end{array}

(Layout) The top three-digit number is 5 A 65\,A\,6 and the two-digit multiplier is 3 B3\,B. The first partial product is 4 C 7 24\,C\,7\,2, the second partial product is 1 D 8 81\,D\,8\,8 written shifted one place to the left, and adding the two partial products gives the final product E 2 0 5 2E\,2\,0\,5\,2.

Show solution

Understand

A three-digit number 5A6 is multiplied by a two-digit number 3B. The two partial products are shown (the second shifted one place left), and the final product is E2052. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 5A6 (a three-digit number)
  • Bottom factor is 3B (a two-digit number)
  • First partial product (top factor times ones digit B) is 4C72
  • Second partial product (top factor times tens digit 3) is 1D88, shifted one place left
  • Final product is E2052
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 3
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 3, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 3. So the top number times 3 must equal 1788. Testing digits, 596 x 3 = 1788, which matches the pattern with A = 9 and D = 7.
596×3=1788⇒A=9, D=7596 \times 3 = 1788 \Rightarrow A=9,\ D=7
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 596 multiplied by the ones digit B and equals 4172. Since 596 x 7 = 4172, we get B = 7 and C = 1.
596×7=4172⇒B=7, C=1596 \times 7 = 4172 \Rightarrow B=7,\ C=1
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 37. Compute 596 x 37 = 4172 + 17880 = 22052, which matches the product with E = 2.
596×37=22052⇒E=2596 \times 37 = 22052 \Rightarrow E=2
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 9, B = 7, C = 1, D = 7, E = 2

Review

Check the whole multiplication: 596 x 37 = 22052, a five-digit number matching the product. Both partial products (4172 and 1788 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (596 x 3 and 596 x 7) and the full product 596 x 37 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 5 answer: A = 3, B = 7, C = 7, D = 1, E = 2

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

5 A 2×4 B3 C 2 42 D 2 80E 5 0 0 4\begin{array}{r} 5\,A\,2 \\ \times \quad 4\,B \\ \hline 3\,C\,2\,4 \\ 2\,D\,2\,8 \phantom{0} \\ \hline E\,5\,0\,0\,4 \end{array}

(Layout) The top three-digit number is 5 A 25\,A\,2 and the two-digit multiplier is 4 B4\,B. The first partial product is 3 C 2 43\,C\,2\,4, the second partial product is 2 D 2 82\,D\,2\,8 written shifted one place to the left, and adding the two partial products gives the final product E 5 0 0 4E\,5\,0\,0\,4.

Show solution

Understand

A three-digit number 5A2 is multiplied by a two-digit number 4B. The two partial products are shown (the second shifted one place left), and the final product is E5004. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 5A2 (a three-digit number)
  • Bottom factor is 4B (a two-digit number)
  • First partial product (top factor times ones digit B) is 3C24
  • Second partial product (top factor times tens digit 4) is 2D28, shifted one place left
  • Final product is E5004
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 4
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 4, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 4. So the top number times 4 must equal 2128. Testing digits, 532 x 4 = 2128, which matches the pattern with A = 3 and D = 1.
532×4=2128⇒A=3, D=1532 \times 4 = 2128 \Rightarrow A=3,\ D=1
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 532 multiplied by the ones digit B and equals 3724. Since 532 x 7 = 3724, we get B = 7 and C = 7.
532×7=3724⇒B=7, C=7532 \times 7 = 3724 \Rightarrow B=7,\ C=7
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 47. Compute 532 x 47 = 3724 + 21280 = 25004, which matches the product with E = 2.
532×47=25004⇒E=2532 \times 47 = 25004 \Rightarrow E=2
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 3, B = 7, C = 7, D = 1, E = 2

Review

Check the whole multiplication: 532 x 47 = 25004, a five-digit number matching the product. Both partial products (3724 and 2128 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (532 x 4 and 532 x 7) and the full product 532 x 47 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 6 answer: A = 4, B = 6, C = 4, D = 4, E = 1

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

7 A 9×2 B4 C 9 41 D 9 80E 9 4 7 4\begin{array}{r} 7\,A\,9 \\ \times \quad 2\,B \\ \hline 4\,C\,9\,4 \\ 1\,D\,9\,8 \phantom{0} \\ \hline E\,9\,4\,7\,4 \end{array}

(Layout) The top three-digit number is 7 A 97\,A\,9 and the two-digit multiplier is 2 B2\,B. The first partial product is 4 C 9 44\,C\,9\,4, the second partial product is 1 D 9 81\,D\,9\,8 written shifted one place to the left, and adding the two partial products gives the final product E 9 4 7 4E\,9\,4\,7\,4.

Show solution

Understand

A three-digit number 7A9 is multiplied by a two-digit number 2B. The two partial products are shown (the second shifted one place left), and the final product is E9474. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 7A9 (a three-digit number)
  • Bottom factor is 2B (a two-digit number)
  • First partial product (top factor times ones digit B) is 4C94
  • Second partial product (top factor times tens digit 2) is 1D98, shifted one place left
  • Final product is E9474
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 2
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 2, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 2. So the top number times 2 must equal 1498. Testing digits, 749 x 2 = 1498, which matches the pattern with A = 4 and D = 4.
749×2=1498⇒A=4, D=4749 \times 2 = 1498 \Rightarrow A=4,\ D=4
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 749 multiplied by the ones digit B and equals 4494. Since 749 x 6 = 4494, we get B = 6 and C = 4.
749×6=4494⇒B=6, C=4749 \times 6 = 4494 \Rightarrow B=6,\ C=4
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 26. Compute 749 x 26 = 4494 + 14980 = 19474, which matches the product with E = 1.
749×26=19474⇒E=1749 \times 26 = 19474 \Rightarrow E=1
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 4, B = 6, C = 4, D = 4, E = 1

Review

Check the whole multiplication: 749 x 26 = 19474, a five-digit number matching the product. Both partial products (4494 and 1498 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (749 x 2 and 749 x 6) and the full product 749 x 26 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 7 answer: A = 2, B = 3, C = 1, D = 8, E = 3

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

7 A 4×4 B2 C 7 22 D 9 60E 1 1 3 2\begin{array}{r} 7\,A\,4 \\ \times \quad 4\,B \\ \hline 2\,C\,7\,2 \\ 2\,D\,9\,6 \phantom{0} \\ \hline E\,1\,1\,3\,2 \end{array}

(Layout) The top three-digit number is 7 A 47\,A\,4 and the two-digit multiplier is 4 B4\,B. The first partial product is 2 C 7 22\,C\,7\,2, the second partial product is 2 D 9 62\,D\,9\,6 written shifted one place to the left, and adding the two partial products gives the final product E 1 1 3 2E\,1\,1\,3\,2.

Show solution

Understand

A three-digit number 7A4 is multiplied by a two-digit number 4B. The two partial products are shown (the second shifted one place left), and the final product is E1132. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 7A4 (a three-digit number)
  • Bottom factor is 4B (a two-digit number)
  • First partial product (top factor times ones digit B) is 2C72
  • Second partial product (top factor times tens digit 4) is 2D96, shifted one place left
  • Final product is E1132
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 4
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 4, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 4. So the top number times 4 must equal 2896. Testing digits, 724 x 4 = 2896, which matches the pattern with A = 2 and D = 8.
724×4=2896⇒A=2, D=8724 \times 4 = 2896 \Rightarrow A=2,\ D=8
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 724 multiplied by the ones digit B and equals 2172. Since 724 x 3 = 2172, we get B = 3 and C = 1.
724×3=2172⇒B=3, C=1724 \times 3 = 2172 \Rightarrow B=3,\ C=1
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 43. Compute 724 x 43 = 2172 + 28960 = 31132, which matches the product with E = 3.
724×43=31132⇒E=3724 \times 43 = 31132 \Rightarrow E=3
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 2, B = 3, C = 1, D = 8, E = 3

Review

Check the whole multiplication: 724 x 43 = 31132, a five-digit number matching the product. Both partial products (2172 and 2896 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (724 x 4 and 724 x 3) and the full product 724 x 43 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 8 answer: A = 8, B = 3, C = 0, D = 3, E = 1

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

6 A 4×2 B2 C 5 21 D 6 80E 5 7 3 2\begin{array}{r} 6\,A\,4 \\ \times \quad 2\,B \\ \hline 2\,C\,5\,2 \\ 1\,D\,6\,8 \phantom{0} \\ \hline E\,5\,7\,3\,2 \end{array}

(Layout) The top three-digit number is 6 A 46\,A\,4 and the two-digit multiplier is 2 B2\,B. The first partial product is 2 C 5 22\,C\,5\,2, the second partial product is 1 D 6 81\,D\,6\,8 written shifted one place to the left, and adding the two partial products gives the final product E 5 7 3 2E\,5\,7\,3\,2.

Show solution

Understand

A three-digit number 6A4 is multiplied by a two-digit number 2B. The two partial products are shown (the second shifted one place left), and the final product is E5732. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 6A4 (a three-digit number)
  • Bottom factor is 2B (a two-digit number)
  • First partial product (top factor times ones digit B) is 2C52
  • Second partial product (top factor times tens digit 2) is 1D68, shifted one place left
  • Final product is E5732
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 2
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 2, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 2. So the top number times 2 must equal 1368. Testing digits, 684 x 2 = 1368, which matches the pattern with A = 8 and D = 3.
684×2=1368⇒A=8, D=3684 \times 2 = 1368 \Rightarrow A=8,\ D=3
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 684 multiplied by the ones digit B and equals 2052. Since 684 x 3 = 2052, we get B = 3 and C = 0.
684×3=2052⇒B=3, C=0684 \times 3 = 2052 \Rightarrow B=3,\ C=0
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 23. Compute 684 x 23 = 2052 + 13680 = 15732, which matches the product with E = 1.
684×23=15732⇒E=1684 \times 23 = 15732 \Rightarrow E=1
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 8, B = 3, C = 0, D = 3, E = 1

Review

Check the whole multiplication: 684 x 23 = 15732, a five-digit number matching the product. Both partial products (2052 and 1368 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (684 x 2 and 684 x 3) and the full product 684 x 23 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 9 answer: A = 6, B = 5, C = 3, D = 3, E = 1

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

4 A 2×3 B2 C 1 01 D 8 60E 6 1 7 0\begin{array}{r} 4\,A\,2 \\ \times \quad 3\,B \\ \hline 2\,C\,1\,0 \\ 1\,D\,8\,6 \phantom{0} \\ \hline E\,6\,1\,7\,0 \end{array}

(Layout) The top three-digit number is 4 A 24\,A\,2 and the two-digit multiplier is 3 B3\,B. The first partial product is 2 C 1 02\,C\,1\,0, the second partial product is 1 D 8 61\,D\,8\,6 written shifted one place to the left, and adding the two partial products gives the final product E 6 1 7 0E\,6\,1\,7\,0.

Show solution

Understand

A three-digit number 4A2 is multiplied by a two-digit number 3B. The two partial products are shown (the second shifted one place left), and the final product is E6170. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 4A2 (a three-digit number)
  • Bottom factor is 3B (a two-digit number)
  • First partial product (top factor times ones digit B) is 2C10
  • Second partial product (top factor times tens digit 3) is 1D86, shifted one place left
  • Final product is E6170
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 3
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 3, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 3. So the top number times 3 must equal 1386. Testing digits, 462 x 3 = 1386, which matches the pattern with A = 6 and D = 3.
462×3=1386⇒A=6, D=3462 \times 3 = 1386 \Rightarrow A=6,\ D=3
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 462 multiplied by the ones digit B and equals 2310. Since 462 x 5 = 2310, we get B = 5 and C = 3.
462×5=2310⇒B=5, C=3462 \times 5 = 2310 \Rightarrow B=5,\ C=3
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 35. Compute 462 x 35 = 2310 + 13860 = 16170, which matches the product with E = 1.
462×35=16170⇒E=1462 \times 35 = 16170 \Rightarrow E=1
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 6, B = 5, C = 3, D = 3, E = 1

Review

Check the whole multiplication: 462 x 35 = 16170, a five-digit number matching the product. Both partial products (2310 and 1386 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (462 x 3 and 462 x 5) and the full product 462 x 35 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 10 answer: A = 6, B = 8, C = 9, D = 8, E = 2

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

3 A 9×5 B2 C 5 21 D 4 50E 1 4 0 2\begin{array}{r} 3\,A\,9 \\ \times \quad 5\,B \\ \hline 2\,C\,5\,2 \\ 1\,D\,4\,5 \phantom{0} \\ \hline E\,1\,4\,0\,2 \end{array}

(Layout) The top three-digit number is 3 A 93\,A\,9 and the two-digit multiplier is 5 B5\,B. The first partial product is 2 C 5 22\,C\,5\,2, the second partial product is 1 D 4 51\,D\,4\,5 written shifted one place to the left, and adding the two partial products gives the final product E 1 4 0 2E\,1\,4\,0\,2.

Show solution

Understand

A three-digit number 3A9 is multiplied by a two-digit number 5B. The two partial products are shown (the second shifted one place left), and the final product is E1402. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 3A9 (a three-digit number)
  • Bottom factor is 5B (a two-digit number)
  • First partial product (top factor times ones digit B) is 2C52
  • Second partial product (top factor times tens digit 5) is 1D45, shifted one place left
  • Final product is E1402
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 5
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 5, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 5. So the top number times 5 must equal 1845. Testing digits, 369 x 5 = 1845, which matches the pattern with A = 6 and D = 8.
369×5=1845⇒A=6, D=8369 \times 5 = 1845 \Rightarrow A=6,\ D=8
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 369 multiplied by the ones digit B and equals 2952. Since 369 x 8 = 2952, we get B = 8 and C = 9.
369×8=2952⇒B=8, C=9369 \times 8 = 2952 \Rightarrow B=8,\ C=9
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 58. Compute 369 x 58 = 2952 + 18450 = 21402, which matches the product with E = 2.
369×58=21402⇒E=2369 \times 58 = 21402 \Rightarrow E=2
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 6, B = 8, C = 9, D = 8, E = 2

Review

Check the whole multiplication: 369 x 58 = 21402, a five-digit number matching the product. Both partial products (2952 and 1845 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (369 x 5 and 369 x 8) and the full product 369 x 58 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!
Variant 11 answer: A = 8, B = 9, C = 2, D = 7, E = 2

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of AA, BB, CC, DD, and EE.

5 A 3×3 B5 C 4 71 D 4 90E 2 7 3 7\begin{array}{r} 5\,A\,3 \\ \times \quad 3\,B \\ \hline 5\,C\,4\,7 \\ 1\,D\,4\,9 \phantom{0} \\ \hline E\,2\,7\,3\,7 \end{array}

(Layout) The top three-digit number is 5 A 35\,A\,3 and the two-digit multiplier is 3 B3\,B. The first partial product is 5 C 4 75\,C\,4\,7, the second partial product is 1 D 4 91\,D\,4\,9 written shifted one place to the left, and adding the two partial products gives the final product E 2 7 3 7E\,2\,7\,3\,7.

Show solution

Understand

A three-digit number 5A3 is multiplied by a two-digit number 3B. The two partial products are shown (the second shifted one place left), and the final product is E2737. We must find each hidden digit A, B, C, D, and E.

Givens
  • Top factor is 5A3 (a three-digit number)
  • Bottom factor is 3B (a two-digit number)
  • First partial product (top factor times ones digit B) is 5C47
  • Second partial product (top factor times tens digit 3) is 1D49, shifted one place left
  • Final product is E2737
Unknowns
  • The digits A, B, C, D, and E
Constraints
  • Each letter is a single digit 0-9
  • The second partial product equals the top number times 3
  • The first partial product equals the top number times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product equals the top number times 3, which depends only on A; testing digits for A pins it down. With A known, the first partial product (top times B) fixes B, and the addition fixes E. Breaking the multiplication into its partial products is the key split.

Execute

#6 Guess and Check 4.NBT.B.5
The second partial product is the top number multiplied by the tens digit 3. So the top number times 3 must equal 1749. Testing digits, 583 x 3 = 1749, which matches the pattern with A = 8 and D = 7.
583×3=1749⇒A=8, D=7583 \times 3 = 1749 \Rightarrow A=8,\ D=7
Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the pattern come out right.
#6 Guess and Check 4.NBT.B.5
The first partial product is 583 multiplied by the ones digit B and equals 5247. Since 583 x 9 = 5247, we get B = 9 and C = 2.
583×9=5247⇒B=9, C=2583 \times 9 = 5247 \Rightarrow B=9,\ C=2
Looking at the ones place narrows B; one multiplication confirms C.
#7 Identify Subproblems 4.NBT.B.5
The full multiplier is 39. Compute 583 x 39 = 5247 + 17490 = 22737, which matches the product with E = 2.
583×39=22737⇒E=2583 \times 39 = 22737 \Rightarrow E=2
Adding the two partial products is multi-digit addition, and the leading digit of the sum gives E.
Answer: A = 8, B = 9, C = 2, D = 7, E = 2

Review

Check the whole multiplication: 583 x 39 = 22737, a five-digit number matching the product. Both partial products (5247 and 1749 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits and read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (583 x 3 and 583 x 9) and the full product 583 x 39 to identify the hidden digits.
💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!