Finding Partial Products (Missing Number) - Solve multiplication equations using partial products

Sep 30, 2026

Uploaded by dhado

Finding Partial Products (Missing Number) provides a set of multiplication problems where students complete missing factors using place value and partial-products reasoning. Each item rewrites a multiplication equation as a sum of partial products, such as splitting by tens and ones, then asks for the number that makes the equation true. An accompanying answer key lists the correct missing numbers for all problems, supporting self-check and instructional use.

LanguageEnglish

CategoryExam

FormatPDF

Pages3

Reading Time8 min

AccessFree to download

Find the number that makes each equation true

Partial products using place value

Answer key

Finding Partial Products (Missing Number) Name: | Find the number that makes each equation true.
1) A student uses place value to break apart a multiplication problem. 597 × 3 =(500 × 3) + (90 × 3) +
2) 62 × 74 = (60 × 70) + (60 × 4) + (2 × 70) +
3) George is using partial products to multiply. 54 × 34 = (50 × 30) + (50 × 4) + (4 ×
30) + (4 × ___)
4) Paige uses place value to break apart a multiplication problem. 154 × 2 = (100 ×
2) + (50 × 2) + (___ × 2)
5) 57 × 32 = (50 × 30) + (50 × 2) + (7 × 30) + (7 × )
6) A teacher shows this partial products strategy on the board. 12 × 46 = (10 × 40)+ (10 × 6) + (2 × 40) + (2 × ___)
7) A student breaks a multiplication problem into partial products. 853 × 2 = (800 × 2) + (50 × 2) + (___ × 2)
8) Haley is using partial products to multiply. 63 × 85 = (60 × 80) + (60 × 5) + (3 ×
80) +
9) 579 × 2 = (500 × 2) + (70 × 2) + ( × 2)
10) 733 × 5 = (700 × 5) + (30 × 5) +
11) A teacher shows this partial products strategy on the board. 54 × 92 = (50 × 90)+ (50 × 2) + (4 × 90) +
12) Oliver breaks a multiplication problem into partial products. 989 × 8 = (900 × 8) +(80 × 8) +
13) 87 × 23 = (80 × 20) + (80 × 3) + (7 × 20) + (7 × )
14) Bianca uses place value to break apart a multiplication problem. 671 × 9 = (600 × 9) + (70 × 9) + (___ × 9)
15) 582 × 6 = (500 × 6) + (80 × 6) + | | | | Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15. | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | |

Math
[www.CommonCoreSheets.com](www.CommonCoreSheets.com)
5
1-10
11-15 | 93 | 87 | 80 | | 73 | 67 | 60 | 53 | 47 | 40 | 33 | | | 27 | 20 | 13 | | 7 | 0 | | | | | | Finding Partial Products (Missing Number) Name: Answer Key | Find the number that makes each equation true.
1) A student uses place value to break apart a multiplication problem. 597 × 3 =(500 × 3) + (90 × 3) + | | | | | Answers
1. 21 | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | 2) | 62 × 74 = (60 × 70) + (60 × 4) + (2 × 70) + | | | | 2. 8 | | | | | | | | | 3) | George is using partial products to multiply. 54 × 34 = (50 × 30) + (50 × 4) + (4 × 30) + (4 × ___) | | | | 3. 4 4. 4 | | | | | | | | | 4) | Paige uses place value to break apart a multiplication problem. 154 × 2 = (100 × 2) + (50 × 2) + (___ × 2) | | | | 5. 2 | | | | | | | | | 5) | 57 × 32 = (50 × 30) + (50 × 2) + (7 × 30) + (7 × ___) | | | | 6. 6 | | | | | | | | | 6) | A teacher shows this partial products strategy on the board. 12 × 46 = (10 × 40)+ (10 × 6) + (2 × 40) + (2 × ___) | | | | 7. 3 8. 15 | | | | | | | | | 7) | A student breaks a multiplication problem into partial products. 853 × 2 = (800 × 2) + (50 × 2) + (___ × 2) | | | | 9. 9 | | | | | | | | | 8) | Haley is using partial products to multiply. 63 × 85 = (60 × 80) + (60 × 5) + (3 × 80) + | | | | 10. 15 | | | | | | | | | 9) | 579 × 2 = (500 × 2) + (70 × 2) + (___ × 2) | | | | 11. 8 | | | | | | | | | 10) | 733 × 5 = (700 × 5) + (30 × 5) + | | | | 12. 72 13. 3 | | | | | | | | | 11) | A teacher shows this partial products strategy on the board. 54 × 92 = (50 × 90)+ (50 × 2) + (4 × 90) + | | | | 14. 1 | | | | | | | | | 12) | Oliver breaks a multiplication problem into partial products. 989 × 8 = (900 × 8) +(80 × 8) + | | | | 15. 12 | | | | | | | | | 13) | 87 × 23 = (80 × 20) + (80 × 3) + (7 × 20) + (7 × ___) | | | | | | | | | | | | | 14) | Bianca uses place value to break apart a multiplication problem. 671 × 9 = (600 × 9) + (70 × 9) + (___ × 9) | | | | | | | | | | | | | 15) | 582 × 6 = (500 × 6) + (80 × 6) + | | | | | | | | | | | | | | Math
[www.Com

FAQs

What skill do the problems test in Finding Partial Products (Missing Number)?

The worksheet tests the ability to use partial products and place value to break apart multiplication and identify the missing number that completes each equation.

How do students solve each multiplication problem?

Students rewrite the multiplication as a sum of partial products (e.g., tens times, ones times), then compute which missing term makes the rewritten equation match the original product.

Is there an answer key included?

Yes. The document includes an answer key listing the correct missing numbers for items 1 through 15.

Sep 30, 2026

Uploaded by dhado