We Distribute, Yet Things Multiply · Lesson 4 of 5
This Way or That Way, All Ways Lead to the Bay
“Explore equivalent counting methods, growing patterns, areas and algebraic applications.”
• Describe growing patterns using more than one algebraic expression. • Prove that different counting and area methods are equivalent. • Translate verbal and visual information into algebra. • Use expansion to investigate calendar, consecutive-number and remainder patterns. • Explain answers with clear assumptions, units and checks.
Imagine counting a large arrangement of counters with a friend. You see a square with one corner missing. Your friend sees a smaller square and two extra rows. You write different expressions, but you are counting the same counters. Neither method needs to be wrong. Algebra can show how the two expressions are connected, and a third arrangement might make the total even easier to calculate.
In this lesson, the important question is not only “What is the answer?” but also “What did you count?” Every term should correspond to a visible group, a length, an area or a clearly described number. Once the expressions are written, expansion lets you check whether the different methods agree.
Start each pattern by identifying its step number and its rule for growing. A few totals can suggest many possible sequences. The arrangement itself supplies the structure that justifies the expression. After deriving a rule, check it on a small stage before using it for a distant one.
Consider a (k + 1) by (k + 1) square arrangement of circles with the bottom-right circle missing. At step 1, there are 3 circles; at step 2, there are 8; at step 3, there are 15. At step 4, the full square would have 25 circles, so the arrangement contains 24.
Count the full square and subtract the missing corner: (k + 1)² − 1. Or count a k by k square, a column of k circles and a row of k circles: k² + k + k. These are two descriptions of the same arrangement.
A third method counts the top k rows, each with k + 1 circles, then adds the remaining row of k circles. It gives k(k + 1) + k. For a fourth method, move those remaining k circles so that each of the top k rows gains one circle. You obtain k rows of k + 2 circles, giving k(k + 2). No circle is added or lost.
| Way of counting | Expression | Expansion |
|---|---|---|
| Full square minus one | (k + 1)² − 1 | k² + 2k + 1 − 1 = k² + 2k |
| Small square and two strips | k² + 2k | k² + 2k |
| Rectangle and remaining row | k(k + 1) + k | k² + k + k = k² + 2k |
| Rearranged rectangle | k(k + 2) | k² + 2k |
Problem
Find the numbers of circles at steps 10 and 15.
- 1.The expression k(k + 2) is convenient because it needs only one multiplication.
- 2.At step 10, substitute k = 10: 10 × 12 = 120. At step 15: 15 × 17 = 255.
- 3.Check step 10 another way: (10 + 1)² − 1 = 121 − 1 = 120. Agreement between methods helps detect substitution errors.
Expressions that have the same value for every allowed value of their letters. They may describe different ways of grouping the same objects. Expanding and collecting like terms is one way to show their equivalence.
How many circles are added from step k to step k + 1 in the missing-corner square pattern? Test your result on steps 2 and 3.
The new count is (k + 1)(k + 3) = k² + 4k + 3. Subtract k² + 2k to obtain 2k + 3. For k = 2, the increase is 7, agreeing with 15 − 8. Geometrically, changing the complete square from side k + 1 to side k + 2 adds a row and a column, with their shared corner counted once; both stages still omit one circle.
Now look at a one-tile-wide square border. At step n, let the empty inner square have side n tiles, so the outside square has side n + 2 tiles. The first three borders contain 8, 12 and 16 tiles.
Subtract the empty inner square from the outside square to obtain (n + 2)² − n². Expanding gives n² + 4n + 4 − n² = 4n + 4. Alternatively, count the top and bottom rows of n + 2 tiles and the left and right sides of n tiles between them. This gives 2(n + 2) + 2n, with no corner counted twice.
Problem
Find the border counts at steps 4 and 10.
- 1.Use the expression 4n + 4. For n = 4, the result is 16 + 4 = 20 tiles.
- 2.For n = 10, the result is 40 + 4 = 44 tiles.
- 3.The outside-minus-inside method confirms the second count: 12² − 10² = 144 − 100 = 44. Always distinguish the inner side n from the outer side n + 2.
Area problems use the same idea of counting a whole and removing known parts. Arrange four identical rectangles of sides m and n around the inside edge of a square of side m + n, with n greater than m. The uncovered centre is a square of side n − m.
One method starts with the outer area (m + n)² and subtracts the four rectangles, each of area mn. Another directly squares the central side n − m. Expanding the first method gives m² + 2mn + n² − 4mn = m² − 2mn + n², exactly the expansion of (n − m)².
Problem
Find the central area when m = 2 and n = 7.
- 1.The outside square has side 9, so its area is 81 square units. Each rectangle has area 14 square units.
- 2.Subtract four rectangles: 81 − 4 × 14 = 81 − 56 = 25 square units.
- 3.The centre has side 7 − 2 = 5, giving 5² = 25 square units directly. Both methods measure the same region.
A second arrangement uses three identical x by y rectangles, with x greater than y: one across the top, one across the bottom, and one standing vertically between them. The shaded region is the space left on either side of the vertical rectangle. Each method below must count those two spaces, without including the rectangles.
Problem
Find the shaded area algebraically, then substitute x = 8 and y = 3.
- 1.Focus on the middle x by x square. Remove its vertical rectangle of area xy: the shaded area is x² − xy.
- 2.Alternatively, use the entire outer rectangle of width x and height x + 2y. Remove all three rectangles: x(x + 2y) − 3xy = x² + 2xy − 3xy = x² − xy.
- 3.A third method counts the two equal shaded rectangles, each of width (x − y)/2 and height x: 2 × x × (x − y)/2 = x(x − y).
- 4.For x = 8 and y = 3, each method gives 64 − 24 = 40 square units. The equal-halves method gives two areas of 8 × 2.5 = 20.
You can also remove a strip from the right of a rectangle and another from its bottom. Let the original width be s, height p and each strip width r, with r smaller than both p and s. The remaining rectangle has sides s − r and p − r.
Direct multiplication gives (p − r)(s − r). Starting with the whole area gives ps − pr − sr + r²: subtract the right strip and bottom strip, then add their r by r overlap once to correct double subtraction. This is the same reasoning used for the square of a difference.
Problem
Find the remaining area when p = 6, s = 9 and r = 3.5.
- 1.The remaining sides are p − r = 2.5 and s − r = 5.5. Multiplying gives 2.5 × 5.5 = 13.75 square units.
- 2.Using removal instead: ps = 54, pr = 21, sr = 31.5 and r² = 12.25.
- 3.Thus 54 − 21 − 31.5 + 12.25 = 13.75 square units. Omitting the overlap correction would give the wrong area.
“Two more than a square number” means s² + 2: square first, then add 2. It differs from (s + 2)², which adds 2 before squaring. “The sum of the squares of two consecutive integers” can be m² + (m + 1)², or m² + (m − 1)² if m is the larger integer. The expression (2m)² + (2m + 1)² describes consecutive pairs whose smaller member is even; it does not describe every pair when m is an integer.
Problem
A two-by-two calendar block has top row a, a + 1 and bottom row a + 7, a + 8. Compare its diagonal products.
- 1.The top-right to bottom-left product is (a + 1)(a + 7) = a² + 8a + 7.
- 2.The top-left to bottom-right product is a(a + 8) = a² + 8a.
- 3.Subtracting gives (a + 1)(a + 7) − a(a + 8) = 7. The first diagonal product is always 7 larger.
- 4.For the block 4, 5 above 11, 12, the products are 55 and 48. Use an actual two-by-two block of consecutive calendar weeks; crossing a row boundary would change the arrangement.
Problem
Take three consecutive integers. Square the middle one and subtract the product of the other two. What remains?
- 1.Write the integers as n − 1, n and n + 1. The required expression is n² − (n − 1)(n + 1).
- 2.Use the difference of squares: (n − 1)(n + 1) = n² − 1.
- 3.Subtract: n² − (n² − 1) = 1. The brackets matter: subtracting −1 adds 1.
- 4.For 5, 6, 7 the result is 36 − 35 = 1. For −2, −1, 0 it is 1 − 0 = 1. The algebra covers negative integers too.
Problem
Which is larger: 14 × 26 or 16 × 24? What about 25 × 75 or 26 × 74?
- 1.The first pair of products has common centre 20: 14 × 26 = 20² − 6², while 16 × 24 = 20² − 4².
- 2.Subtracting the smaller square leaves the larger result, so 16 × 24 is larger by 6² − 4² = 20.
- 3.The second pair has centre 50: 25 × 75 = 50² − 25² and 26 × 74 = 50² − 24².
- 4.Therefore 26 × 74 is larger by 25² − 24² = (25 + 24)(25 − 24) = 49. With a fixed sum, these closer factors give the larger product.
Problem
Prove that an even integer has a square divisible by 4, and an odd integer has a square that is 1 more than a multiple of 8.
- 1.Write an even integer as 2t. Its square is 4t², which is 4 times an integer.
- 2.Write an odd integer as 2t + 1. Its square is 4t² + 4t + 1 = 4t(t + 1) + 1.
- 3.The consecutive integers t and t + 1 include an even member, so t(t + 1) = 2u for an integer u.
- 4.The odd square is therefore 8u + 1. This is a proof for every integer t, including values that make 2t + 1 negative.
For integer variables, decide whether each is always true: (k + 1)(k + 2) − (k + 3) is 2; (2q + 1)(2q − 3) is divisible by 4; (6n + 2)² − (4n + 3)² is 5 less than an integer square.
All three universal claims are false. The first expands to k² + 2k − 1; k = 0 gives −1, not 2. The second expands to 4q² − 4q − 3 = 4(q² − q − 1) + 1, so it leaves remainder 1, not 0. The third expands to 20n² − 5. If it were 5 less than a square for every n, then 20n² would always be a square; n = 1 gives 20, which lies strictly between 4² and 5². A valid counterexample is enough to refute an “always” claim.
Problem
A leaves remainder 3 when divided by 7, and B leaves remainder 5. Find the remainders of A + B, A − B, B − A and AB.
- 1.Write A = 7u + 3 and B = 7v + 5. Their sum is 7(u + v) + 8 = 7(u + v + 1) + 1, so its remainder is 1.
- 2.A − B = 7(u − v) − 2 = 7(u − v − 1) + 5. Its standard non-negative remainder is 5.
- 3.B − A = 7(v − u) + 2, so its remainder is 2. The subtraction order changes the answer.
- 4.AB = 49uv + 35u + 21v + 15 = 7(7uv + 5u + 3v + 2) + 1. Its remainder is 1.
- 5.If “difference” means the positive difference, you must know which original number is larger. Remainders alone do not tell you that.
Add two numbers and multiply the result by half of that sum. Write and simplify an expression.
Let the numbers be a and b. The sum is a + b, and half of it is (a + b)/2. Their product is (a + b)(a + b)/2 = (a + b)²/2. Expanding gives a²/2 + ab + b²/2. For a = 4 and b = 2, the product is 6 × 3 = 18, which is half of 36.
For a park with two square lawns, an area subtraction can save a long count of individual paving tiles. Let each lawn have side g feet. In the layout below the outer margins are w feet wide and the central gap is 2w feet wide, because it includes a w-wide margin beside each lawn. These labelled widths determine the answer.
The outside width is w + g + 2w + g + w = 2g + 4w, and the outside height is w + g + w = g + 2w. Subtract both lawn areas g² from the total rectangle. Keep square feet as the area unit; a width in feet is not itself an area.
Problem
Find a simplified expression for the walking area, then evaluate it for g = 10 ft and w = 2 ft.
- 1.Multiply the outside dimensions: (2g + 4w)(g + 2w) = 2g² + 4gw + 4gw + 8w².
- 2.Subtract 2g² for the lawns. The walking area is 8gw + 8w² = 8w(g + w) square feet.
- 3.For g = 10 and w = 2, this is 8 × 2 × 12 = 192 square feet.
- 4.Check with dimensions: the outside is 28 ft by 14 ft, with area 392 square feet. Remove 200 square feet of lawn to get 192. A layout with a central gap of only w would have different dimensions and a different answer.
Two final tile patterns show why a diagram can explain a sequence more clearly than its first few totals. In the yellow pattern at step y, a central rectangle has y rows and y + 2 columns. One vertical arm of y + 2 tiles extends above its right edge, and another of y + 2 tiles extends below its left edge. The arms do not overlap the central rectangle.
Its count is y(y + 2) + 2(y + 2) = (y + 2)(y + 2) = (y + 2)². In the blue pattern at step y, there is a (y + 1) by (y + 1) square and y additional tiles beneath it. Its count is (y + 1)² + y = y² + 3y + 1.
Problem
Find the counts at steps 4 and 10 in both final tile patterns.
- 1.Yellow pattern: at y = 4, the count is (4 + 2)² = 36; at y = 10, it is 12² = 144. Draw a 4 by 6 centre with two arms of 6 tiles for the next stage.
- 2.Blue pattern: at y = 4, the count is 5² + 4 = 29; at y = 10, it is 11² + 10 = 131. Draw a 5 by 5 square with 4 extra tiles for the next stage.
- 3.Checking the first three stages gives 9, 16, 25 and 5, 11, 19 respectively. The decomposition explains the rules beyond those initial checks.
Quiz
Which expression counts a (k + 1)-by-(k + 1) square of circles with one missing corner?
Why is a one-tile border around an n-by-n hole counted by 2(n + 2) + 2n?
In a calendar block, which diagonal product is 7 larger?
A has remainder 3 and B remainder 5 modulo 7. What is the standard remainder of A − B?
Two positive factors keep the same sum. Which change increases their product?
Answer: k² + 2k. The count is (k + 1)² − 1 = k² + 2k. The constant +1 and the missing circle cancel.
Practice Problems
- Find the circle count at step 12 using two equivalent expressions.
- The inner side of a one-tile-wide square border is 8 tiles. Find the border count and explain how corners are handled.
- For the three-rectangle shaded-area arrangement, use x = 10 and y = 4. Find the shaded area by all three methods.
- A calendar block starts with 12 in its top-left corner. Find all four entries and prove the diagonal difference without relying on this single example.
- Two integers A and B leave remainders 3 and 5 on division by 7. Find the remainders of A + B, AB and A − B. Then explain why the positive difference is not determined by the remainders alone.
Using k(k + 2) gives 12 × 14 = 168. Using (k + 1)² − 1 gives 13² − 1 = 169 − 1 = 168. Both expressions expand to k² + 2k, so their agreement is general.
Key Takeaways
• A diagram can support several valid counting methods; explain what each term counts. • Expanding and collecting terms proves that different expressions are equivalent. • In a border or cutout, count corners and overlapping regions exactly once. • Pattern formulas allow distant-stage predictions when the growing structure is specified. • Calendar and consecutive-number relationships can be proved using general variables. • The subtraction order matters when finding remainders of a difference. • Use the labelled dimensions of an area problem and include square units in the answer.