Chapter 4 · Sections 2 & 3Textbook pages 65–73

Number Properties & Curiosities ക്രിയാവിശേഷങ്ങൾ · ക്രിയാകൗതുകങ്ങൾ

Checking a few examples can make you suspect that something is always true. Algebra lets you prove it. In this part of the chapter the identity (x + y)(u + v) = xu + xv + yu + yv proves facts about odd numbers and remainders, and then explains three number tricks: reversed digits, calendar squares and multiplication-table squares.

📖 What you already know

(x + y)(u + v) = xu + xv + yu + yv – every term of the first sum times every term of the second (page 5).
General forms: even numbers 2n, odd numbers 2n + 1, numbers leaving remainder 1 on division by 3 are 3n + 1, two-digit numbers are 10m + n (page 2).
An identity (സർവസമവാക്യം) is true for every value of the letters, so one algebraic calculation covers infinitely many examples.

1. From examples to proof

7 × 9 = 63, 11 × 13 = 143, 25 × 31 = 775 – odd times odd gave an odd number every time. But "every time I tried" is not "always". The textbook asks: how can we prove this for all odd numbers? The answer is to write a general odd number with a letter and multiply with the identity.

2. Odd × odd is always odd

🎬 Proof animation: (2m + 1)(2n + 1)

🎚️ Try any m and n

m n

Whatever m and n you choose, the product is 2 × (something) + 1 – an odd number. The algebra guarantees it; the slider only illustrates it.

⭐ Key fact (textbook page 65)

(2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1. The product of two odd numbers is one more than an even number, so it is odd. രണ്ട് ഒറ്റസംഖ്യകളുടെ ഗുണനഫലവും ഒറ്റസംഖ്യ തന്നെയാണ്.

3. Remainders: numbers of the form 3n + 1

Odd numbers leave remainder 1 when divided by 2. What about numbers that leave remainder 1 when divided by 3 – 1, 4, 7, 10, 13, …? They are all 3n + 1 (മൂന്നിന്റെ ഗുണിതത്തിനോട് 1 കൂട്ടിയവ). Multiply two of them: 4 × 7 = 28 = 3 × 9 + 1, 10 × 13 = 130 = 3 × 43 + 1. Remainder 1 again. Proof:

🎬 Proof animation: (3m + 1)(3n + 1)

🧪 Remainder laboratory – any divisor, any remainders

divisor d remainder r remainder s

m n

(dm + r)(dn + s) = d(dmn + ms + nr) + rs, so the product leaves the same remainder as r × s does. This lab previews Exercise 2.

4. Four consecutive numbers

Take 1, 2, 3, 4: the ends multiply to 1 × 4 = 4, the middle pair to 2 × 3 = 6. Take 2, 3, 4, 5: 2 × 5 = 10 and 3 × 4 = 12. The difference is 2 both times. Always?

🎚️ Slide along the number line

first number x

🎬 Proof animation: x(x + 3) and (x + 1)(x + 2)

⭐ Key fact (textbook page 66)

അടുത്തടുത്തുള്ള ഏതു നാല് എണ്ണൽസംഖ്യകളെടുത്താലും, അറ്റത്തുള്ള സംഖ്യകളുടെ ഗുണനഫലവും, നടുക്കുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലവും തമ്മിലുള്ള വ്യത്യാസം രണ്ട് ആണ്.
For any four consecutive natural numbers, (product of the middle two) − (product of the ends) = 2, because (x + 1)(x + 2) = x² + 3x + 2 and x(x + 3) = x² + 3x.

5. Curiosity 1: reverse the digits, keep the product

12 × 63 = 756 and 21 × 36 = 756. Also 32 × 46 = 1472 and 23 × 64 = 1472. Reverse both numbers and the product does not change! Usually that fails: 12 × 34 = 408 but 21 × 43 = 903. So when does it work?

🎬 Finding the rule with 10m + n

🎚️ Test any two two-digit numbers

m (tens of the 1st) n (units of the 1st)

p (tens of the 2nd) q (units of the 2nd)

The rule: reversing both numbers keeps the product exactly when m × p = n × q (tens digit × tens digit = units digit × units digit). Pairs where the second number is just the first reversed, like 12 × 21, are left out of the list because they are not interesting.

6. Curiosity 2: the calendar square

In any month, mark a 2 × 2 square of dates, for example 6, 7, 13, 14. Multiply diagonally: 14 × 6 = 84 and 13 × 7 = 91. Difference 7. Try 22, 23, 29, 30: 22 × 30 = 660, 23 × 29 = 667. Difference 7 again. Click any square below.

🖐️ Click a date to choose a 2 × 2 square

Blue cells are one diagonal, amber cells the other. The difference of the two diagonal products is 7 for every square in every month.

🎬 Why always 7? Call the top-left date x

7. Curiosity 3: squares in the multiplication table

Here the trick uses sums, not products. In the table of 1 to 9, take any 2 × 2 square, say 12, 16, 15, 20. Add diagonally: 12 + 20 = 32 and 16 + 15 = 31. Difference 1. For 35, 40, 42, 48: 35 + 48 = 83 and 40 + 42 = 82. Click any square.

🖐️ Click a cell of the table (ഗുണിതപ്പട്ടിക)

Row n contains n, 2n, 3n, …, 9n. The square beginning at the m-th entry of row n holds mn, (m + 1)n, m(n + 1) and (m + 1)(n + 1).

🎬 Why always 1? Write the four numbers with m and n

🌍 Real life: tricks and checks

Magicians use these facts: ask a friend to circle any 2 × 2 square on a calendar and tell you only the top-left date; you instantly announce both diagonal products (x(x + 8) and that number plus 7). Bank clerks use a similar idea for a quick check: if a total is off by a multiple of 9, two digits were probably swapped – because 10m + n and 10n + m differ by 9(m − n). Algebra turns a curiosity into a tool.

8. Check yourself

🧭 Summary of Sections 2 & 3

To prove a fact about all numbers of a kind, write the general form (2n + 1, 3n + 1, 10m + n, x, x + 1, …) and multiply with the identity.
(2m + 1)(2n + 1) = 2(2mn + m + n) + 1: odd × odd is odd. (3m + 1)(3n + 1) = 3(3mn + m + n) + 1: remainder 1 stays remainder 1.
Four consecutive numbers: (x + 1)(x + 2) − x(x + 3) = 2.
Reversing both two-digit numbers keeps the product exactly when mp = nq (12 × 63 = 21 × 36).
Calendar 2 × 2 square: (x + 1)(x + 7) − x(x + 8) = 7. Multiplication-table 2 × 2 square: diagonal sums differ by 1.

Next: Exercises 2 & 3 – all questions solved with animations ▶