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The document includes an “Answer Key” section listing the correct missing numbers for each problem.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},463866,1791310878,{"code":4,"msg":82,"data":83},"success",[84,88,92,95,99,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & 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teacher shows this partial products strategy on the board. 53 × 52 = (50 × 50)+ (50 × 2) + (3 × 50) + (3 × ___)\u003Cbr>2) 18 × 18 = (10 × 10) + (10 × 8) + (8 × 10) +  \u003Cbr>3) A student uses place value to break apart a multiplication problem. 312 × 8 =(300 × 8) + (10 × 8) +  \u003Cbr>4) 473 × 8 = (400 × 8) + (70 × 8) +  \u003Cbr>5) A student breaks a multiplication problem into partial products. 678 × 8 = (600 × 8) + (70 × 8) + (___ × 8)\u003Cbr>6) A teacher shows this partial products strategy on the board. 95 × 16 = (90 × 10)+ (90 × 6) + (5 × 10) +  \u003Cbr>7) Jerry is using partial products to multiply. 55 × 14 = (50 × 10) + (50 × 4) + (5 × 10) + (5 × ___)\u003Cbr>8) Amy is using partial products to multiply. 91 × 84 = (90 × 80) + (90 × 4) + (1 × 80)+  \u003Cbr>9) Emily uses place value to break apart a multiplication problem. 571 × 2 = (500 ×\u003Cbr>2) + (70 × 2) + (___ × 2)\u003Cbr>10) Paul breaks a multiplication problem into partial products. 227 × 8 = (200 × 8) +(20 × 8) +  \u003Cbr>11) 561 × 2 = (500 × 2) + (60 × 2) + (   × 2)\u003Cbr>12) 84 × 34 = (80 × 30) + (80 × 4) + (4 × 30) + (4 ×   )\u003Cbr>13) 718 × 7 = (700 × 7) + (10 × 7) +  \u003Cbr>14) A student breaks a multiplication problem into partial products. 233 × 3 = (200 × 3) + (30 × 3) + (___ × 3)\u003Cbr>15) A teacher shows this partial products strategy on the board. 98 × 33 = (90 × 30)+ (90 × 3) + (8 × 30) +   |  |  |  | Answers\u003Cbr>1.  \u003Cbr>2.  \u003Cbr>3.  \u003Cbr>4.  \u003Cbr>5.  \u003Cbr>6.  \u003Cbr>7.  \u003Cbr>8.  \u003Cbr>9.  \u003Cbr>10.  \u003Cbr>11.  \u003Cbr>12.  \u003Cbr>13.  \u003Cbr>14.  \u003Cbr>15.   |  |  |  |  |  |  |  |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| \u003Cbr>\u003Cbr>Math\u003Cbr>[www.CommonCoreSheets.com](www.CommonCoreSheets.com)\u003Cbr>1\u003Cbr>1-10\u003Cbr>11-15 | 93 | 87 | 80 |  | 73 | 67 | 60 | 53 | 47 | 40 | 33 |\n|  | 27 | 20 | 13 |  | 7 | 0 |  |  |  |  |  |\n\nFinding Partial Products (Missing Number)  \nName: Answer Key  \n\n| Find the number that makes each equation true.\u003Cbr>1) A teacher shows this partial products strategy on the board. 53 × 52 = (50 × 50)+ (50 × 2) + (3 × 50) + (3 × ___) |  |  |  |  | Answers\u003Cbr>1.  2  |  |  |  |  |  |  |  |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 2) | 18 × 18 = (10 × 10) + (10 × 8) + (8 × 10) +   |  |  |  | 2.  64  |  |  |  |  |  |  |  |\n| 3) | A student uses place value to break apart a multiplication problem. 312 × 8 =(300 × 8) + (10 × 8) +   |  |  |  | 3.  16  4. 24 |  |  |  |  |  |  |  |\n| 4) | 473 × 8 = (400 × 8) + (70 × 8) +   |  |  |  | 5.  8  |  |  |  |  |  |  |  |\n| 5) | A student breaks a multiplication problem into partial products. 678 × 8 = (600 × 8) + (70 × 8) + (___ × 8) |  |  |  | 6.  30  |  |  |  |  |  |  |  |\n| 6) | A teacher shows this partial products strategy on the board. 95 × 16 = (90 × 10)+ (90 × 6) + (5 × 10) +   |  |  |  | 7.  4  8. 4 |  |  |  |  |  |  |  |\n| 7) | Jerry is using partial products to multiply. 55 × 14 = (50 × 10) + (50 × 4) + (5 × 10) + (5 × ___) |  |  |  | 9.  1  |  |  |  |  |  |  |  |\n| 8) | Amy is using partial products to multiply. 91 × 84 = (90 × 80) + (90 × 4) + (1 × 80)+   |  |  |  | 10.  56  |  |  |  |  |  |  |  |\n| 9) | Emily uses place value to break apart a multiplication problem. 571 × 2 = (500 × 2) + (70 × 2) + (___ × 2) |  |  |  | 11.  1  |  |  |  |  |  |  |  |\n| 10) | Paul breaks a multiplication problem into partial products. 227 × 8 = (200 × 8) +(20 × 8) +   |  |  |  | 12.  4  13. 56 |  |  |  |  |  |  |  |\n| 11) | 561 × 2 = (500 × 2) + (60 × 2) + (   × 2) |  |  |  | 14.  3  |  |  |  |  |  |  |  |\n| 12) | 84 × 34 = (80 × 30) + (80 × 4) + (4 × 30) + (4 × ___) |  |  |  | 15.  24  |  |  |  |  |  |  |  |\n| 13) | 718 × 7 = (700 × 7) + (10 × 7) +   |  |  |  |  |  |  |  |  |  |  |  |\n| 14) | A student breaks a multiplication problem into partial products. 233 × 3 = (200 × 3) + (30 × 3) + (___ × 3) |  |  |  |  |  |  |  |  |  |  |  |\n| 15) | A teacher shows this partial products strategy on the board. 98 × 33 = (90","cbCaiqKp19g9Jzl9","https://ap.wps.com/l/cbCaiqKp19g9Jzl9","pdf",43614,"English","# Find the number that makes each equation true\n## Partial products strategy examples\n## Place value decomposition examples\n## Answer key","[{\"question\":\"What skills do the problems practice?\",\"answer\":\"The worksheet practices computing missing numbers in multiplication using partial products and place value decomposition.\"},{\"question\":\"How do students solve each equation?\",\"answer\":\"They break the multiplication into expanded partial products (tens, ones, and remainders) and determine the missing factor that completes the expression.\"},{\"question\":\"Is an answer key provided?\",\"answer\":\"Yes. The document includes an “Answer Key” section listing the correct missing numbers for each problem.\"}]","Finding Partial Products (Missing Number) | PDF",1790767287]