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An accompanying answer key provides the correct missing numbers, covering examples with two- and three-digit multiplicands and single-digit multipliers, using strategies such as tens and ones decomposition.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/finding-partial-products-missing-number-answers/485418/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/finding-partial-products-missing-number-answers/485418.png","ImageObject",300,407,{"name":42,"@type":43},"dhado","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-11-06","2026-09-30",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",7,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"What skill do these problems practice?","Question",{"text":63,"@type":64},"They practice finding a missing number in multiplication equations by using partial products and place value decomposition.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How do students approach each multiplication problem?",{"text":68,"@type":64},"Students rewrite the multiplicand into expanded form (tens, hundreds, etc.) and add the partial products, then use the structure to find the missing factor.",{"name":70,"@type":61,"acceptedAnswer":71},"Is an answer key included?",{"text":72,"@type":64},"Yes. The document includes an answer section listing the missing numbers for each problem from 1 to 15.","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},485418,1792789676,{"code":4,"msg":82,"data":83},"success",[84,88,92,95,100,105,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & 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true.\u003Cbr>1) Tom is using partial products to multiply. 45 × 56 = (40 × 50) + (40 × 6) + (5 × 50)+ (5 × ___)\u003Cbr>2) 41 × 67 = (40 × 60) + (40 × 7) + (1 × 60) + (1 ×   )\u003Cbr>3) A student uses place value to break apart a multiplication problem. 162 × 2 =(100 × 2) + (60 × 2) +  \u003Cbr>4) Isabel is using partial products to multiply. 35 × 61 = (30 × 60) + (30 × 1) + (5 ×\u003Cbr>60) +  \u003Cbr>5) A teacher shows this partial products strategy on the board. 94 × 93 = (90 × 90)+ (90 × 3) + (4 × 90) +  \u003Cbr>6) A student breaks a multiplication problem into partial products. 123 × 5 = (100 × 5) + (20 × 5) + (___ × 5)\u003Cbr>7) Amy uses place value to break apart a multiplication problem. 766 × 7 = (700 × 7) + (60 × 7) + (___ × 7)\u003Cbr>8) A teacher shows this partial products strategy on the board. 28 × 94 = (20 × 90)+ (20 × 4) + (8 × 90) + (8 × ___)\u003Cbr>9) Luke breaks a multiplication problem into partial products. 695 × 9 = (600 × 9) +(90 × 9) +  \u003Cbr>10) 78 × 43 = (70 × 40) + (70 × 3) + (8 × 40) +  \u003Cbr>11) 337 × 4 = (300 × 4) + (30 × 4) + (   × 4)\u003Cbr>12) 724 × 3 = (700 × 3) + (20 × 3) +  \u003Cbr>13) 175 × 4 = (100 × 4) + (70 × 4) + (   × 4)\u003Cbr>14) A student breaks a multiplication problem into partial products. 971 × 8 = (900 × 8) + (70 × 8) + (___ × 8)\u003Cbr>15) 282 × 5 = (200 × 5) + (80 × 5) +   |  |  |  | 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>10\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) Tom is using partial products to multiply. 45 × 56 = (40 × 50) + (40 × 6) + (5 × 50)+ (5 × ___) |  |  |  |  | Answers\u003Cbr>1.  6  |  |  |  |  |  |  |  |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 2) | 41 × 67 = (40 × 60) + (40 × 7) + (1 × 60) + (1 ×   ) |  |  |  | 2.  7  |  |  |  |  |  |  |  |\n| 3) | A student uses place value to break apart a multiplication problem. 162 × 2 =(100 × 2) + (60 × 2) +   |  |  |  | 3.  4  4. 5 |  |  |  |  |  |  |  |\n| 4) | Isabel is using partial products to multiply. 35 × 61 = (30 × 60) + (30 × 1) + (5 × 60) +   |  |  |  | 5.  12  |  |  |  |  |  |  |  |\n| 5) | A teacher shows this partial products strategy on the board. 94 × 93 = (90 × 90)+ (90 × 3) + (4 × 90) +   |  |  |  | 6.  3  |  |  |  |  |  |  |  |\n| 6) | A student breaks a multiplication problem into partial products. 123 × 5 = (100 × 5) + (20 × 5) + (___ × 5) |  |  |  | 7.  6  8. 4 |  |  |  |  |  |  |  |\n| 7) | Amy uses place value to break apart a multiplication problem. 766 × 7 = (700 × 7) + (60 × 7) + (___ × 7) |  |  |  | 9.  45  |  |  |  |  |  |  |  |\n| 8) | A teacher shows this partial products strategy on the board. 28 × 94 = (20 × 90)+ (20 × 4) + (8 × 90) + (8 × ___) |  |  |  | 10.  24  |  |  |  |  |  |  |  |\n| 9) | Luke breaks a multiplication problem into partial products. 695 × 9 = (600 × 9) +(90 × 9) +   |  |  |  | 11.  7  12. 12 |  |  |  |  |  |  |  |\n| 10) | 78 × 43 = (70 × 40) + (70 × 3) + (8 × 40) +   |  |  |  | 13. 5 |  |  |  |  |  |  |  |\n| 11) | 337 × 4 = (300 × 4) + (30 × 4) + (___ × 4) |  |  |  | 14.  1  |  |  |  |  |  |  |  |\n| 12) | 724 × 3 = (700 × 3) + (20 × 3) +   |  |  |  | 15.  10  |  |  |  |  |  |  |  |\n| 13) | 175 × 4 = (100 × 4) + (70 × 4) + (___ × 4) |  |  |  |  |  |  |  |  |  |  |  |\n| 14) | A student breaks a multiplication problem into partial products. 971 × 8 = (900 × 8) + (70 × 8) + (___ × 8) |  |  |  |  |  |  |  |  |  |  |  |\n| 15) | 282 × 5 = (200 × 5) + (80 × 5) +   |  |  |  |  |  |  |  |  |  |  |  |\n|  | Math\u003Cbr>[www.CommonCoreSheets.com](www.CommonCoreSheets.com)","cbCaieAEwluFfpHe","https://ap.wps.com/l/cbCaieAEwluFfpHe","pdf",43635,"English","# Finding Partial Products (Missing Number) - Equations\n## Problem Set (1-15)\n# Answer Key","[{\"question\":\"What skill do these problems practice?\",\"answer\":\"They practice finding a missing number in multiplication equations by using partial products and place value decomposition.\"},{\"question\":\"How do students approach each multiplication problem?\",\"answer\":\"Students rewrite the multiplicand into expanded form (tens, hundreds, etc.) and add the partial products, then use the structure to find the missing factor.\"},{\"question\":\"Is an answer key included?\",\"answer\":\"Yes. The document includes an answer section listing the missing numbers for each problem from 1 to 15.\"}]","Finding Partial Products (Missing Number) - Answers | PDF",1790804184]