FREE 1-Digit Partial Product Multiplication Worksheet - 4th Grade ... - Free Printable

The task involves solving multiplication problems using the partial product method. The partial product method breaks down a multiplication problem into smaller, more manageable parts by multiplying each digit of one number by each digit of the other number, considering their place values. Let's solve each problem step by step.

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Problem 1: \( 617 \times 5 \)

#### Step 1: Break down \( 617 \) into its place values.
- \( 617 = 600 + 10 + 7 \)

#### Step 2: Multiply each part by 5.
- \( 600 \times 5 = 3000 \)
- \( 10 \times 5 = 50 \)
- \( 7 \times 5 = 35 \)

#### Step 3: Add the partial products.
- \( 3000 + 50 + 35 = 3085 \)

#### Final Answer:
\[ \boxed{3085} \]

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Problem 2: \( 23 \times 8 \)

#### Step 1: Break down \( 23 \) into its place values.
- \( 23 = 20 + 3 \)

#### Step 2: Multiply each part by 8.
- \( 20 \times 8 = 160 \)
- \( 3 \times 8 = 24 \)

#### Step 3: Add the partial products.
- \( 160 + 24 = 184 \)

#### Final Answer:
\[ \boxed{184} \]

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Problem 3: \( 825 \times 6 \)

#### Step 1: Break down \( 825 \) into its place values.
- \( 825 = 800 + 20 + 5 \)

#### Step 2: Multiply each part by 6.
- \( 800 \times 6 = 4800 \)
- \( 20 \times 6 = 120 \)
- \( 5 \times 6 = 30 \)

#### Step 3: Add the partial products.
- \( 4800 + 120 + 30 = 4950 \)

#### Final Answer:
\[ \boxed{4950} \]

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Problem 4: \( 137 \times 3 \)

#### Step 1: Break down \( 137 \) into its place values.
- \( 137 = 100 + 30 + 7 \)

#### Step 2: Multiply each part by 3.
- \( 100 \times 3 = 300 \)
- \( 30 \times 3 = 90 \)
- \( 7 \times 3 = 21 \)

#### Step 3: Add the partial products.
- \( 300 + 90 + 21 = 411 \)

#### Final Answer:
\[ \boxed{411} \]

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Problem 5: \( 764 \times 4 \)

#### Step 1: Break down \( 764 \) into its place values.
- \( 764 = 700 + 60 + 4 \)

#### Step 2: Multiply each part by 4.
- \( 700 \times 4 = 2800 \)
- \( 60 \times 4 = 240 \)
- \( 4 \times 4 = 16 \)

#### Step 3: Add the partial products.
- \( 2800 + 240 + 16 = 3056 \)

#### Final Answer:
\[ \boxed{3056} \]

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Problem 6: \( 239 \times 9 \)

#### Step 1: Break down \( 239 \) into its place values.
- \( 239 = 200 + 30 + 9 \)

#### Step 2: Multiply each part by 9.
- \( 200 \times 9 = 1800 \)
- \( 30 \times 9 = 270 \)
- \( 9 \times 9 = 81 \)

#### Step 3: Add the partial products.
- \( 1800 + 270 + 81 = 2151 \)

#### Final Answer:
\[ \boxed{2151} \]

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Problem 7: \( 231 \times 7 \)

#### Step 1: Break down \( 231 \) into its place values.
- \( 231 = 200 + 30 + 1 \)

#### Step 2: Multiply each part by 7.
- \( 200 \times 7 = 1400 \)
- \( 30 \times 7 = 210 \)
- \( 1 \times 7 = 7 \)

#### Step 3: Add the partial products.
- \( 1400 + 210 + 7 = 1617 \)

#### Final Answer:
\[ \boxed{1617} \]

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Problem 8: \( 953 \times 8 \)

#### Step 1: Break down \( 953 \) into its place values.
- \( 953 = 900 + 50 + 3 \)

#### Step 2: Multiply each part by 8.
- \( 900 \times 8 = 7200 \)
- \( 50 \times 8 = 400 \)
- \( 3 \times 8 = 24 \)

#### Step 3: Add the partial products.
- \( 7200 + 400 + 24 = 7624 \)

#### Final Answer:
\[ \boxed{7624} \]

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Problem 9: \( 337 \times 2 \)

#### Step 1: Break down \( 337 \) into its place values.
- \( 337 = 300 + 30 + 7 \)

#### Step 2: Multiply each part by 2.
- \( 300 \times 2 = 600 \)
- \( 30 \times 2 = 60 \)
- \( 7 \times 2 = 14 \)

#### Step 3: Add the partial products.
- \( 600 + 60 + 14 = 674 \)

#### Final Answer:
\[ \boxed{674} \]

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Final Answers for All Problems:

1. \( \boxed{3085} \)
2. \( \boxed{184} \)
3. \( \boxed{4950} \)
4. \( \boxed{411} \)
5. \( \boxed{3056} \)
6. \( \boxed{2151} \)
7. \( \boxed{1617} \)
8. \( \boxed{7624} \)
9. \( \boxed{674} \)

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Explanation of the Partial Product Method:

The partial product method simplifies multiplication by breaking numbers into their place values (e.g., hundreds, tens, ones) and multiplying each part separately. This approach helps in understanding the mechanics of multiplication and is particularly useful for multi-digit numbers. By adding the results of these smaller multiplications, we arrive at the final product. This method reinforces place value concepts and builds a strong foundation in arithmetic.