Models are Fun
Draw an area model to show 23 × 14. Label each rectangle with its numerical value and write the total product.
Draw 20×10, 3×10, 20×4, 3×4 → 200 + 30 + 80 + 12 = 322.
Use the distributive property to compute 12 × 15 by breaking 15 into 10 + 5. Show each step and the final product.
12×(10+5)=120+60=180
Use the standard algorithm to compute 24 × 36. Show all steps and the final product.
24×36 = 864.
Use the associative property to show how to compute (12 × 5) × 3 and 12 × (5 × 3). Compute both and state whether they give the same result.
(12×5)×3=60×3=180; 12×(5×3)=12×15=180.
Use the partial-products method to compute 32 × 14. List each partial product and sum them.
32×14: 30×10=300,30×4=120,2×10=20,2×4=8 → total 448.
Two rectangles are shown in an area model: one partition is 20 × 34 and the other is 3 × 34. Complete the area model and explain how it represents 23 × 34. Calculate the product.
20×34 = 680; 3×34 = 102; total = 782; explains 23×34.
Write two different distributive decompositions for 37 × 26 (for example, decompose 37 or 26 differently). Compute both and explain why both give the same product.
Two valid decompositions e.g., 37×(20+6)=740+222=962; (30+7)×26=780+182=962
the standard algorithm to 45 × 28 but forgot to shift the second partial product (write it under the wrong place value). Show the correct arrangement of partial products and explain the error and correction.
Correct layout shows 45×28: (45×8=360) and (45×20=900 shift one place) → 1260; explanation of shifting
Use the associative property to simplify computing 25 × 48 by regrouping factors (for example, pair with a factor that makes a tens or hundreds). Show your regrouping and compute the product.
Example regrouping: 25×48 = 25×(4×12) etc.; one computed result 1200.
For 58 × 23, use partial products to compute. Then rearrange the partial products to show a connection to the distributive property.
58×23: 50×20=1000,50×3=150,8×20=160,8×3=24 → 1334; relate to distributive sums.
A student splits 46 × 57 into (40 + 6) × 57 and draws the area model. Explain why the area model confirms the result of multiplying 46 × 57 and compute the product.
(40×57=2280)+(6×57=342)=2622
A student computes 49 × 36 by doing 49 × (30 + 6). Show the full distributive steps, then explain how this method connects to the standard algorithm.
49×30=1470; 49×6=294; total 1764; connects to place-value.
Given 67 × 54, perform the multiplication using the standard algorithm. Then, explain step-by-step how each line of the standard algorithm corresponds to place-value reasoning.
67×54: 67×4=268; 67×50=3350; sum=3618; relate to place value of each row.
Given the expression 4 × (27 × 13), use the associative property to rearrange the multiplication into a sequence that makes mental computation easier. Compute the result and explain your choice of grouping.
4×(27×13) → (4×27)×13 =108×13=1404 (or 4×(13×27)=4×351=1404).
A student uses partial products for 74 × 36 and records: 70×30, 70×6, 4×30, 4×6. Compute each and sum. Then explain why partial products always match the standard algorithm.
74×36: 70×30=2100,70×6=420,4×30=120,4×6=24 →2644.
Design an area-model strategy to compute 78 × 64 by choosing an efficient partition (not necessarily tens and ones). Show your partition, the partial rectangle values, and the sum. Explain why your partition reduces computation time.
Example: partition 78 as 80−2 or 70+8; one correct answer: 78=(70+8), 64=(60+4): 70×60=4200,70×4=280,8×60=480,8×4=32 → 4992; explain efficiency.
Create a word problem that requires using the distributive property to multiply 63 × 47 where mental math or partitioning makes the problem easier. Solve your problem and justify why the distributive approach is efficient.
Example word problem and solution; justification depends on chosen numbers.
Compare the standard algorithm result and the partial products result for 89 × 63. Show both methods, confirm they match, and write a short justification that the standard algorithm is a compact form of the partial products method.
89×63 partial products: 89×60=5340; 89×3=267; sum=5607; show equivalence.
A problem gives 15 × 26 × 4. Use associative reasoning to find the most efficient order to compute this product mentally. Show two different groupings, evaluate both, and explain which is faster and why.
15×26×4 group (15×4)=60; 60×26=1560 (efficient); other grouping yields same product.
Given a complex two-digit multiplication 96 × 87, use partial products. Show an efficient ordering of the partial products to make addition easier and explain your reasoning.
96×87 partials: 90×80=7200,90×7=630,6×80=480,6×7=42 → 8352; order them to add easily.
Given a rectangle of area 2,976 with side lengths whole numbers where one side is between 70 and 80 and the other is between 30 and 40, use area reasoning to determine the exact two-digit factors. Show how this relates to computing a two-digit multiplication problem.
Factors: 2,976 = 78 × 38 (show checking) → relates to 78×38.
Prove that for any two-digit numbers using the distributive property leads to the same product as the standard algorithm. Use a specific example (e.g., 84 × 79) and show algebraic and numeric work.
Algebraic demonstration: (10a+b)(10c+d)=100ac+10(ad+bc)+bd; numeric example e.g., 84×79= (80+4)(70+9)=5600+720+280+36=6636 matches standard algorithm.
You are asked to teach the standard algorithm for multiplying two two-digit numbers to a small group. Design a 5-step lesson outline that builds conceptual understanding (not just procedure), include one formative assessment question, and demonstrate the lesson with 76 × 48.
Lesson outline and worked example: 76×48 → 76×8=608; 76×40=3040; total 3648.
Create a real-world problem (multi-step) where students must apply the associative property to multiply three numbers (each two digits) to find an answer. Solve it, showing how grouping affects ease of computation but not the final result.
Example multi-step problem and full solution; final numeric check.
Provide a multi-part task: (a) Compute 83 × 59 using partial products. (b) Estimate the product by rounding to nearest ten and compare the estimate to the exact answer. (c) Reflect: which method (partial products or estimate-first) gives a better check for reasonableness and why?
(a) 83×59 partials: 80×50=4000,80×9=720,3×50=150,3×9=27 → 4897. (b) Rounded: 80×60=4800 estimate close. (c) Partial products give exact check; estimate gives reasonableness.