Chapter 4 · Exercises 2 & 3Textbook pages 66–67 and 73

Exercises 2 & 3 – Animated Solutions ക്രിയാവിശേഷങ്ങൾ · ക്രിയാകൗതുകങ്ങൾ

Six questions about proving number patterns. The recipe is always the same: check with numbers, guess the rule, then prove it with letters using (x + y)(u + v) = xu + xv + yu + yv. Try each question in your notebook first, then watch the solution.

📖 Tools you will use

General forms (page 2): remainder r on division by d → dn + r; two-digit number → 10m + n; consecutive numbers → x, x + 1, x + 2, …
Four-piece expansion (page 5) and the proofs of Sections 2–3 (page 7).
A multiple of d plus r leaves remainder r (if r < d).

Exercise 2 · Question 1 – check, guess, prove

Q1Check each of the following with several numbers, guess a general rule, and prove it with algebra:

(ചുവടെപ്പറയുന്ന ഓരോ കാര്യവും, പല സംഖ്യകളെടുത്ത് പരിശോധിക്കുക. അവയിൽനിന്ന് പൊതുവായ ഒരു തത്വം ഊഹിക്കുക. ഊഹിച്ചത് ശരിയാണെന്ന് ബീജഗണിതം ഉപയോഗിച്ച് തെളിയിക്കുക.)

The remainder on dividing by 3 the product of a number leaving remainder 1 and a number leaving remainder 2 (on division by 3).
The remainder on dividing by 4 the product of a number leaving remainder 1 and a number leaving remainder 2 (on division by 4).
For six consecutive natural numbers, the difference between the product of the two end numbers and the product of the two middle numbers.

(i) The numbers are 3m + 1 and 3n + 2. (ii) They are 4m + 1 and 4n + 2. (iii) Call the six numbers x, x + 1, …, x + 5: the ends are x and x + 5, the middle two are x + 2 and x + 3. Expand both products and compare.

(i) Divisor 3, remainders 1 and 2

🧪 Check with numbers

m n

🎬 Proof: (3m + 1)(3n + 2)

(i) (3m + 1)(3n + 2) = 9mn + 6m + 3n + 2 = 3(3mn + 2m + n) + 2 ⇒ the remainder is always 2. Checks: 4 × 5 = 20 = 3 × 6 + 2; 7 × 8 = 56 = 3 × 18 + 2.

(ii) Divisor 4, remainders 1 and 2

🧪 Check with numbers

m n

🎬 Proof: (4m + 1)(4n + 2)

(ii) (4m + 1)(4n + 2) = 16mn + 8m + 4n + 2 = 4(4mn + 2m + n) + 2 ⇒ the remainder is always 2. Checks: 5 × 6 = 30 = 4 × 7 + 2; 9 × 10 = 90 = 4 × 22 + 2.

(iii) Six consecutive numbers

🧪 Check with numbers

first number x

🎬 Proof: x(x + 5) and (x + 2)(x + 3)

(iii) Ends: x(x + 5) = x² + 5x. Middles: (x + 2)(x + 3) = x² + 5x + 6. The middle product is larger by 6, always. (1…6: 12 − 6 = 6; 2…7: 20 − 14 = 6.)

💡 Pattern of patterns

Four consecutive numbers gave difference 2 = 1 × 2; six give 6 = 2 × 3. Eight would give (x + 3)(x + 4) − x(x + 7) = 12 = 3 × 4. The difference is always (half the count − 1) × (half the count).

Exercise 2 · Question 2 – the 36 × 28 scheme

Q2A way of finding 36 × 28 is shown below. (i) Try this method with other two-digit numbers. (ii) Explain, using algebra, why it works. (Remember from Class 7 that any two-digit number can be written as 10m + n.)

3 × 2 = 6 6 × 100 600 (3 × 8) + (6 × 2) = 36 36 × 10 360 6 × 8 = 48 48 ───────────── 36 × 28 1008

Write 36 = 10 × 3 + 6 and 28 = 10 × 2 + 8, and expand (10m + n)(10p + q) into four pieces. Group the two pieces that have a factor 10.

🧮 (i) Try it with any two two-digit numbers

first number second number

Three rows: (tens × tens) hundreds, (cross products added) tens, (units × units) ones. The picture shows the four pieces; the two green pieces (units × tens and tens × units) are the middle row, the red corner is units × units.

🎬 (ii) Why it works: (10m + n)(10p + q)

(ii) (10m + n)(10p + q) = 100mp + 10mq + 10np + nq = 100 × mp + 10 × (mq + np) + nq. The three rows of the scheme are exactly these three groups: 100 × (3 × 2), 10 × (3 × 8 + 6 × 2), and 6 × 8. (i) e.g. 47 × 53: 20 → 2000, 4 × 3 + 7 × 5 = 27 → 270, 21 → 2491 ✓.

Exercise 3 · Question 1 – a 5-column grid

Q1Write the numbers 1 to 25 in five rows of five, as in the figure. (i) Mark a square of four numbers as in the calendar, multiply diagonally and find the difference. Is the difference the same for every such square? (ii) Explain why, using algebra.

In this grid the number below x is x + 5 (not x + 7 as in a calendar). So the square is x, x + 1, x + 5, x + 6.

🖐️ Click a number to choose a 2 × 2 square

Blue cells are one diagonal, amber the other. The amber product is always 5 more.

🎬 (ii) Why always 5?

(i) e.g. 7, 8, 12, 13: 8 × 12 − 7 × 13 = 96 − 91 = 5; 17, 18, 22, 23: 18 × 22 − 17 × 23 = 396 − 391 = 5. Yes, always 5. (ii) Square x, x + 1, x + 5, x + 6: (x + 1)(x + 5) − x(x + 6) = (x² + 6x + 5) − (x² + 6x) = 5. In a calendar the row length is 7, here it is 5 – the difference equals the row length.

Exercise 3 · Question 2 – a 3 × 3 square in the multiplication table

Q2Instead of a square of four numbers, take a square of nine numbers in the multiplication table and mark only the four corner numbers (the figure shows 6, 8, 10 / 9, 12, 15 / 12, 16, 20 with corners 6, 10, 12, 20). (i) What is the difference between the diagonal sums of the corners? (ii) Explain with algebra why every such square gives difference 4. (iii) What if we take a square of sixteen numbers?

The corners of a 3 × 3 square starting at mn are mn, m(n + 2), (m + 2)n and (m + 2)(n + 2). Expand the last one. For 4 × 4 replace 2 by 3.

🎚️ Any square, any size

row n column m size k

Red cells are the corners. For a k × k square the diagonal sums of the corners differ by (k − 1)²: 1, 4, 9, 16 for k = 2, 3, 4, 5.

🎬 (ii) and (iii): the algebra

(i) 6 + 20 = 26 and 10 + 12 = 22: difference 4. (ii) Corners mn, m(n + 2), (m + 2)n, (m + 2)(n + 2): one diagonal sums to mn + mn + 2m + 2n + 4 = 2mn + 2m + 2n + 4, the other to mn + 2m + mn + 2n = 2mn + 2m + 2n. Difference 4 for every m, n. (iii) Sixteen numbers (4 × 4): corners mn, m(n + 3), (m + 3)n, (m + 3)(n + 3) ⇒ difference 9. In general (k − 1)².

🌍 Why this matters

"Check, guess, prove" is how all of mathematics – and much of science – works. Computer scientists use the same remainder arithmetic (they call it modular arithmetic) to check credit-card numbers and to encrypt messages: the rule "remainder of a product = product of remainders" that you proved here is the engine behind it.

Check yourself

🧭 Summary of Exercises 2 & 3

(dm + r)(dn + s) = d(dmn + ms + nr) + rs: the product leaves the remainder of r × s. Remainders 1 and 2 give 2 (for d = 3 and d = 4).
Six consecutive numbers: (x + 2)(x + 3) − x(x + 5) = 6.
(10m + n)(10p + q) = 100mp + 10(mq + np) + nq – the three-row scheme.
Grid with 5 columns: (x + 1)(x + 5) − x(x + 6) = 5. Multiplication table, k × k square: corner diagonal sums differ by (k − 1)².

Next: Sections 4 & 5 – multiplying differences ▶