Arithmetic Expressions · Lesson 5 of 7
The Distributive Property and Smarter Multiplication
“See how repeated groups explain bracket expansion, common factors and multiplication near convenient numbers.”
• Derive distribution over addition and subtraction using repeated groups. • Multiply every term of a bracketed sum or difference by its factor. • Read distribution in reverse to collect a common factor. • Predict a changed product from a known product. • Use convenient nearby numbers to multiply mentally. • Find and compare different expressions for number-grid totals.
Removing Brackets When Multiplying
Adding or subtracting a bracketed quantity was about performing one combined gain or loss. Multiplying a bracket repeats the whole quantity several times. Each part inside therefore appears in every repeated group, which gives a new reason for removing brackets.
Problem
Lhamo and Norbu each buy a cutlet for ₹43 and a rasgulla for ₹24. Write the combined cost in two ways.
- 1.One person’s order costs 43 + 24 = 67. Two whole orders cost 2 × (43 + 24) = 134.
- 2.Alternatively, count two cutlets and two rasgullas separately: 2 × 43 + 2 × 24 = 86 + 48 = 134.
- 3.Thus 2 × (43 + 24) = 2 × 43 + 2 × 24. The factor 2 applies to both parts of the order.
- 4.For three friends buying the same order, 3 × (43 + 24) = 3 × 43 + 3 × 24 = 129 + 72 = ₹201.
2 × (43 + 24) is not 2 × 43 + 24. The latter counts two cutlets but only one rasgulla and gives ₹110. Ask how many times each item actually appears in the repeated whole.
The Distributive Property
The snack orders illustrate multiplication spreading across an addition. The same reasoning works whenever a number multiplies a sum: each part is repeated that many times. Reading the relationship in reverse lets you combine two products that have a common factor.
Multiplying a sum by a number gives the same value as multiplying each addend by that number and adding the products. Multiplying a difference works similarly, keeping the subtraction between the products.
Problem
There are four rows of scouts and three rows of guides, with five people in every row. Count all the people in two ways.
- 1.The scouts contribute 4 × 5 = 20 and the guides 3 × 5 = 15, giving 4 × 5 + 3 × 5 = 35.
- 2.All seven rows have the same row size. Combine the row counts first: (4 + 3) × 5 = 7 × 5 = 35.
- 3.Therefore 4 × 5 + 3 × 5 = (4 + 3) × 5. The shared factor 5 records the common number of people per row.
Compare 5 × 4 + 3 with 5 × (4 + 3). The first has five groups of four and just three extra objects, giving 23. The second has five groups each containing seven, giving 35. Within the second expression, adding 4 and 3 can be swapped, and the two factors can also be swapped: 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5. Swapping the factors of a product is the commutative property of multiplication; it does not permit changing a sum into a different product.
Problem
Explain 10 × 98 + 3 × 98 without separately multiplying and adding.
- 1.The first product represents ten copies of 98; the second represents three more copies.
- 2.Together there are thirteen copies, so 10 × 98 + 3 × 98 = (10 + 3) × 98.
- 3.Equivalently, reverse the factors: 98 × 10 + 98 × 3 = 98 × (10 + 3).
- 4.The value is 13 × 98 = 1274. Both expanded products must use the same group size, 98.
Distributing Over Subtraction
A difference can also count groups. If you start with fourteen groups of ten and remove six groups of ten, eight complete groups remain. The subtraction concerns whole equal-sized groups, so the same factor belongs to both products.
Problem
Derive a product form for 14 × 10 − 6 × 10.
- 1.Start with fourteen equal groups of ten. Removing six groups leaves 14 − 6 = 8 groups.
- 2.The remaining total is (14 − 6) × 10 = 8 × 10 = 80.
- 3.Thus 14 × 10 − 6 × 10 = (14 − 6) × 10. Expanding the bracket reverses this reasoning.
Write the general relationships below only after seeing the repeated groups. The letters a, b and c stand for numbers. In a × (b + c), a is the factor multiplying both inner terms; in (a + b) × c, c is the common factor. The versions with a minus sign work in the same way, preserving the subtraction. These are equalities that can be read in either direction.
Tinker the Terms II
You can use a known product to find another product nearby. Decide how many equal groups have been added or removed, calculate their contribution, and adjust the known value. Distribution supplies the exact reason the shortcut works.
Problem
Given 53 × 18 = 954, find 63 × 18 without multiplying from the beginning.
- 1.The number of groups increases from 53 to 63, an increase of 10. Each extra group contains 18.
- 2.Write (53 + 10) × 18 = 53 × 18 + 10 × 18.
- 3.Use the known product: 954 + 180 = 1134. Increasing the first factor by 10 increases the product by 10 × 18, not just by 10.
Multiply Near a Convenient Number
Numbers close to 10, 50 or 100 often suggest an easy bracketed sum or difference. Multiply the convenient part first and then account for the small adjustment. Choose a nearby number that makes the products genuinely easier to compute.
Problem
Calculate 97 × 25 using 100 as the nearby number.
- 1.Since 97 = 100 − 3, write 97 × 25 = (100 − 3) × 25.
- 2.Distribute: 100 × 25 − 3 × 25 = 2500 − 75.
- 3.The result is 2425. Three groups of 25 were removed from one hundred groups.
| Product | Convenient rewrite | Value |
|---|---|---|
| 95 × 8 | (100 − 5) × 8 = 800 − 40 | 760 |
| 104 × 15 | (100 + 4) × 15 = 1500 + 60 | 1560 |
| 49 × 50 | (50 − 1) × 50 = 2500 − 50 | 2450 |
Try 98 × 7, 103 × 9 and 49 × 5. The nearby numbers 100, 100 and 50 suggest 700 − 14 = 686, 900 + 27 = 927 and 250 − 5 = 245. Explain why each correction is a product, then propose another multiplication that benefits from the same idea.
Number Grids and Different Counting Routes
In a number grid, the total can be found by rows, columns or counts of equal entries. Each route describes the same set of numbers. Look for repeated rows or repeated values so that the distributive property can simplify the expression.
Problem
Find each grid’s sum using repeated values and using repeated rows.
- 1.Grid I has five 4s and four 8s: 5 × 4 + 4 × 8 = 20 + 32 = 52.
- 2.Its top and bottom rows match. By rows: 2 × (4 + 8 + 4) + (8 + 4 + 8) = 2 × 16 + 20 = 52.
- 3.Grid II has eight 5s and eight 6s: 8 × 5 + 8 × 6 = 8 × (5 + 6) = 88.
- 4.Every row of Grid II totals 22, so 4 × (5 + 6 + 6 + 5) = 4 × 22 = 88. Different grouping gives the same sum.
Check Your Understanding
Try each question before opening the explanations. A useful answer includes the reason for your choice, not just its letter.
Quiz
Which equals 7 × (9 + 2)?
Which common-factor form equals 6 × 13 + 4 × 13?
What is 99 × 16 using a nearby hundred?
If 42 × 12 = 504, what is 45 × 12?
Compare (34 − 28) × 42 with 34 × 42 − 28 × 42.
Seven copies of the entire sum contain seven copies of both 9 and 2.
Work these problems on paper and explain any rewrite you use. Open the matching solution after you have made a first attempt; more than one valid expression or method may be possible.
Practice Problems
- Complete distribution for all sixteen patterns: 3 × (6 + 7); (8 + 3) × 4; 3 × (5 + 8); (9 + 2) × 4; 3 × (□ + 4); (13 + 6) × 4; 3 × (5 + 2); (2 + 3) × 4; 5 × (9 − 2); (5 − 2) × 7; 5 × (8 − 3); (8 − 3) × 7; 5 × (12 − □); (15 − 6) × 7; 5 × (9 − 4); (17 − 9) × 7. Choose a number for each free blank, then use it consistently.
- Compare without fully evaluating: (8 − 3) × 29 and (3 − 8) × 29; 15 + 9 × 18 and (15 + 9) × 18; 23 × (17 − 9) and 23 × 17 + 23 × 9; (34 − 28) × 42 and 34 × 42 − 28 × 42.
- Find five expressions of the form a × (b + c) with value 14, using positive whole numbers.
- Use a convenient nearby number to calculate 95 × 8, 104 × 15 and 49 × 50.
- Given 53 × 18 = 954, find 52 × 18 and 53 × 20 without starting again.
- Write at least two different expressions for the total of each displayed grid and verify they agree.
- A student writes 8 × (20 − 3) = 8 × 20 − 3. Correct the expression and explain the error.
In order: 3 × 6 + 3 × 7; 8 × 4 + 3 × 4; 3 × 5 + 3 × 8; 9 × 4 + 2 × 4; choosing 10 gives 3 × 10 + 3 × 4; 13 × 4 + 6 × 4; 3 × 5 + 3 × 2; 2 × 4 + 3 × 4; 5 × 9 − 5 × 2; 5 × 7 − 2 × 7; 5 × 8 − 5 × 3; 8 × 7 − 3 × 7; choosing 3 gives 5 × 12 − 5 × 3; 15 × 7 − 6 × 7; 5 × 9 − 5 × 4; 17 × 7 − 9 × 7. The two free blanks have many valid choices; the chosen number must match its expanded product.
Key Takeaways
• A factor multiplying a bracket multiplies every inner term. • Distribution works over both addition and subtraction. • Collecting a common factor is distribution read in reverse. • Extra or removed groups change a product by the group count times its size. • Convenient nearby numbers support exact mental multiplication when adjustments are included. • Rows, columns and equal-entry counts can give different expressions for the same grid total.