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📐 Chapter 5 · NCERT 2025-26

🎲 Number Play

Divisibility Rules · Palindromes · Magic Squares · Cryptarithmetic · Number Patterns

1729
121
3×3
SEND
9801
📐 Introduction to Number Play

Numbers are not just tools for calculation — they are full of surprises, patterns, and magic! In this chapter, we explore the playful side of mathematics. From clever divisibility tests that tell you whether a number divides evenly, to palindromic numbers that read the same forwards and backwards, to ancient magic squares and modern cryptarithmetic puzzles, you will discover that numbers can be as entertaining as any game.

The NCERT Ganita textbook for Class 8 invites you to play with numbers, investigate patterns, make conjectures, and verify them. This chapter is about thinking like a mathematician — asking "why does this happen?" and "will this always work?"

🇮🇳 Indian Heritage — A Love for Numbers

India has a deep and ancient tradition of number play. Indian mathematicians were among the first to explore divisibility, digit patterns, and number magic.

📜 Vedic Mathematics

Ancient Indian texts contain Sutras (formulas) for lightning-fast divisibility checks and mental arithmetic. The Vedic approach treats numbers as living, playful entities with hidden patterns.

🌟 Ramanujan (1887–1920)

Srinivasa Ramanujan, the great Indian genius, had an extraordinary intuition for numbers. The famous 1729 (Hardy-Ramanujan number) is the smallest number expressible as the sum of two cubes in two different ways: 1³ + 12³ = 9³ + 10³ = 1729.

💫 Aryabhata (476 CE)

Aryabhata explored divisibility and remainders in his Aryabhatiya. His work on modular arithmetic laid foundations for the divisibility rules we study today.

🧮 Narayana Pandit (1356 CE)

Indian mathematician Narayana Pandit studied magic squares extensively. He devised systematic methods to construct magic squares of any size — centuries before Europeans explored the same topic.
💡 Chapter Goals: By the end of this chapter you will be able to: (1) test divisibility by 2, 3, 4, 5, 6, 8, 9, 10, and 11 without dividing; (2) identify and create palindromic numbers; (3) solve magic square puzzles; (4) decode cryptarithmetic puzzles; (5) discover and explain number patterns.
📚 What You Will Learn
Topic Key Concepts
Divisibility Rules Tests for 2, 3, 4, 5, 6, 8, 9, 10, 11
Palindromic Numbers Numbers that read the same forwards and backwards
Number Patterns Digit sum patterns, reverse-and-add, Kaprekar routine
Magic Squares Squares where rows, columns, diagonals have equal sums
Cryptarithmetic Puzzles where digits replace letters in arithmetic
Famous Numbers Hardy-Ramanujan, Fermat, Armstrong, Kaprekar numbers
🔢 Divisibility Rules

A number a is divisible by another number b if dividing a by b leaves a remainder of zero. Instead of performing long division, we can use quick divisibility tests based on the digits of the number.

🔢 Complete Divisibility Rules Table
Divisor Rule Example
2 Last digit is 0, 2, 4, 6, or 8 (even) 4,536 → last digit 6 (even) → ✅ divisible by 2
3 Sum of digits is divisible by 3 621 → 6+2+1 = 9 → 9÷3 = 3 ✅
4 Last two digits form a number divisible by 4 3,724 → last two digits 24 → 24÷4 = 6 ✅
5 Last digit is 0 or 5 8,475 → last digit 5 ✅
6 Divisible by both 2 and 3 312 → even ✅ & 3+1+2 = 6 (div by 3) ✅
8 Last three digits form a number divisible by 8 5,016 → last three 016 = 16 → 16÷8 = 2 ✅
9 Sum of digits is divisible by 9 2,709 → 2+7+0+9 = 18 → 18÷9 = 2 ✅
10 Last digit is 0 4,530 → last digit 0 ✅
11 Alternating sum of digits (odd − even positions) is 0 or divisible by 11 61,853 → (6+8+3)−(1+5) = 17−6 = 11 ✅
💡 Understanding the Rule for 11

The divisibility rule for 11 is the trickiest. Starting from the leftmost digit, alternate between adding and subtracting each digit. If the result is 0 or a multiple of 11, the original number is divisible by 11.

Example: Is 918,082 divisible by 11?

Step 1: Write the digits: 9, 1, 8, 0, 8, 2
Step 2: Alternating sum from left: 9 − 1 + 8 − 0 + 8 − 2 = 22
Step 3: Is 22 divisible by 11? Yes! 22 ÷ 11 = 2. ✅
Answer: 918,082 is divisible by 11.
💡 Why Do These Rules Work?

Divisibility rules are not arbitrary tricks — they arise from the place value system. For example, why does the digit sum rule work for 9? Consider 234 = 2×100 + 3×10 + 4. Since 100 = 99+1 and 10 = 9+1, we get: 234 = 2×(99+1) + 3×(9+1) + 4 = (2×99 + 3×9) + (2+3+4). The terms 2×99 + 3×9 are always divisible by 9, so 234 is divisible by 9 if and only if the digit sum 2+3+4 = 9 is divisible by 9!

💡 Memory Aid for Divisibility:
• 2 → Last digit Even
• 3 → digit Sum ÷ 3
• 4 → last Two digits ÷ 4
• 5 → Last digit 0 or 5
• 6 → 2 AND 3
• 8 → last Three digits ÷ 8
• 9 → digit Sum ÷ 9
• 10 → last digit 0
• 11 → Alternating sum ÷ 11

Example: Test 7,425,600 for divisibility by 2, 3, 4, 5, 6, 8, 9, 10, 11.

By 2: Last digit = 0 (even) ✅
By 3: Digit sum = 7+4+2+5+6+0+0 = 24. 24÷3 = 8 ✅
By 4: Last two digits = 00. 0÷4 = 0 ✅
By 5: Last digit = 0 ✅
By 6: Divisible by both 2 and 3 ✅
By 8: Last three digits = 600. 600÷8 = 75 ✅
By 9: Digit sum = 24. 24÷9 = 2 remainder 6 ❌ Not divisible by 9
By 10: Last digit = 0 ✅
By 11: 7−4+2−5+6−0+0 = 6. 6 is not divisible by 11 ❌
💡 Important: To test for divisibility by 6, you must check both the rule for 2 AND the rule for 3. A number that is even but whose digit sum is not divisible by 3 (like 14) is NOT divisible by 6.
🔁 Palindromic Numbers

A palindromic number (or simply palindrome) is a number that reads the same forwards and backwards. The word "palindrome" comes from the Greek words palin (again) and dromos (path) — literally "running back again."

🔢 Single-digit Palindromes

All single-digit numbers (0–9) are palindromes: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

🔢 Two-digit Palindromes

11, 22, 33, 44, 55, 66, 77, 88, 99 — there are exactly 9 of them.

🔢 Three-digit Palindromes

Examples: 121, 131, 252, 343, 505, 747, 999. The first and last digits must match. There are 90 three-digit palindromes (from 101 to 999).

🔢 Larger Palindromes

1221, 12321, 45654, 1234321 — palindromes can be as large as you want!
🔁 The Reverse-and-Add Process

Here is a fascinating property: take almost any number, reverse its digits, and add the two numbers. Repeat this process, and most numbers eventually become palindromes! This is called the reverse-and-add process.

Example: Start with 56. Apply reverse-and-add until you reach a palindrome.

Step 1: 56 + 65 = 121 ✅ Palindrome reached in 1 step!

Example: Start with 78. Apply reverse-and-add.

Step 1: 78 + 87 = 165 (not a palindrome)
Step 2: 165 + 561 = 726 (not a palindrome)
Step 3: 726 + 627 = 1353 (not a palindrome)
Step 4: 1353 + 3531 = 4884 ✅ Palindrome reached in 4 steps!
💡 Fun Fact: The number 196 is the smallest number for which no one has ever found a palindrome through reverse-and-add, even after billions of iterations on supercomputers! It is called a Lychrel number candidate. Nobody has proved that 196 will never become a palindrome — it is one of the unsolved mysteries of mathematics!
🔢 Properties of Palindromic Numbers
  • Every palindrome with an even number of digits is divisible by 11. Example: 1221 → 1−2+2−1 = 0 ✅
  • A three-digit palindrome aba equals 100a + 10b + a = 101a + 10b. Since 101 is prime and 10 is not divisible by 101, the structure constrains what values work.
  • The product of a palindrome with its reverse is the palindrome squared: the reverse of a palindrome is itself!
💡 Palindrome in Words: Just like "MADAM" and "RACECAR" are word palindromes, numbers like 12321 are number palindromes. The NCERT textbook encourages you to investigate how many palindromes exist between any two given numbers.
💡 Patterns in Numbers
🔢 Digit Sum Patterns

The digit sum (or digital root) of a number is found by repeatedly summing the digits until you get a single digit. This reveals beautiful hidden patterns.

Example: Find the digital root of 9,875.

Step 1: 9 + 8 + 7 + 5 = 29
Step 2: 2 + 9 = 11
Step 3: 1 + 1 = 2. The digital root of 9,875 is 2.

🎲 Pattern: Multiples of 9

The digital root of every multiple of 9 is always 9!
9 → 9, 18 → 9, 27 → 9, 36 → 9, 45 → 9, ..., 99 → 9, 108 → 9

🎲 Pattern: Squares

The digital roots of perfect squares cycle: 1, 4, 9, 7, 7, 9, 4, 1, 9 and then repeat! A number whose digital root is 2, 3, 5, 6, or 8 cannot be a perfect square.
🔢 The Kaprekar Routine (6174)

Take any 4-digit number (with at least two different digits). Arrange its digits in descending order and ascending order. Subtract the smaller from the larger. Repeat. You will always reach 6174 (called Kaprekar's constant) within 7 steps!

Example: Start with 3,087.

Step 1: Descending: 8730, Ascending: 0378 → 8730 − 0378 = 8352
Step 2: 8532 − 2358 = 6174 ✅ Kaprekar's constant reached in 2 steps!

Example: Start with 2,005.

Step 1: 5200 − 0025 = 5175
Step 2: 7551 − 1557 = 5994
Step 3: 9954 − 4599 = 5355
Step 4: 5553 − 3555 = 1998
Step 5: 9981 − 1899 = 8082
Step 6: 8820 − 0288 = 8532
Step 7: 8532 − 2358 = 6174 ✅
💡 Did You Know? D.R. Kaprekar (1905–1986) was an Indian mathematician from Devlali, Maharashtra. He discovered this remarkable constant and many other number curiosities. Despite being a schoolteacher with no formal research position, his contributions to recreational mathematics are celebrated worldwide.
🔢 Interesting Number Patterns

💡 1 × 1 = 1

11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
Pattern: Pyramid of digits!

💡 9 × 1 + 1 = 10

9 × 12 + 3 = 111
9 × 123 + 4 = 1111
9 × 1234 + 5 = 11111
Pattern: Repunit magic!

💡 1 × 8 + 1 = 9

12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
Pattern: Descending digits!

💡 1 × 9 + 2 = 11

12 × 9 + 3 = 111
123 × 9 + 4 = 1111
1234 × 9 + 5 = 11111
Pattern: Repunits again!
🔢 Armstrong Numbers

An Armstrong number (or narcissistic number) is a number that equals the sum of its own digits each raised to the power of the number of digits. For example:

  • 153 = 1³ + 5³ + 3³ = 1 + 125 + 27 = 153 ✅
  • 370 = 3³ + 7³ + 0³ = 27 + 343 + 0 = 370 ✅
  • 371 = 3³ + 7³ + 1³ = 27 + 343 + 1 = 371 ✅
  • 407 = 4³ + 0³ + 7³ = 64 + 0 + 343 = 407 ✅
💡 Puzzle Challenge: There are exactly four 3-digit Armstrong numbers: 153, 370, 371, and 407. Can you verify each one? For 2-digit numbers, there are none! For 1-digit numbers, all single digits 0–9 are trivially Armstrong numbers.
✨ Magic Squares

A magic square is a grid of numbers where every row, every column, and both main diagonals add up to the same sum, called the magic constant.

✨ The 3×3 Magic Square

The most famous magic square uses the numbers 1 through 9 and has a magic constant of 15.

Magic Constant for n×n square using 1 to n² = n(n² + 1) / 2 For a 3×3 square: 3(9+1)/2 = 3×10/2 = 15
✨ The Classic 3×3 Magic Square
Every row, column, and diagonal adds up to 15!
2
7
6
9
5
1
4
3
8

Row 1: 2+7+6=15 • Row 2: 9+5+1=15 • Row 3: 4+3+8=15
Col 1: 2+9+4=15 • Col 2: 7+5+3=15 • Col 3: 6+1+8=15
Diag: 2+5+8=15 • Diag: 6+5+4=15

🛠 How to Build a 3×3 Magic Square

The NCERT textbook teaches the Siamese method (also called the "staircase method") for odd-order magic squares:

  1. Place 1 in the middle cell of the top row.
  2. Move diagonally up-right to place the next number.
  3. If you go above the top row, wrap to the bottom of the same column.
  4. If you go past the right edge, wrap to the left of the same row.
  5. If the target cell is already occupied, place the number directly below the current cell instead.

🇮🇳 Indian Heritage

A 4×4 magic square appears in the Khajuraho temple (10th century CE) in Madhya Pradesh! It uses the numbers 1–16 with a magic constant of 34. Narayana Pandit (1356 CE) gave systematic methods for constructing magic squares of any order.

🔢 Properties

In the 3×3 magic square with 1–9, the centre cell is always 5 (the middle number). The four corner numbers always add up to 20. The magic constant for an n×n square using 1 to n² is always n(n²+1)/2.
✨ The 4×4 Magic Square

A 4×4 magic square using numbers 1 to 16 has a magic constant of 34. The famous Khajuraho magic square is:

🇮🇳 Khajuraho Temple Magic Square
7
12
1
14
2
13
8
11
16
3
10
5
9
6
15
4

Every row, column, and diagonal sums to 34. Check it yourself!

💡 Fun Activity: Can you find other properties of the 4×4 magic square? Try adding the four corner numbers (7+14+9+4 = 34!). Try adding the four centre numbers (13+8+3+10 = 34!). There are many hidden patterns in magic squares beyond rows, columns, and diagonals.
🔐 Cryptarithmetic Puzzles

Cryptarithmetic (also called alphametic or verbal arithmetic) is a type of puzzle where digits are replaced by letters. Each letter stands for a unique digit (0–9), and the leading digit of a number cannot be 0. Your job is to figure out which digit each letter represents so that the arithmetic is correct.

🔐 The Classic: SEND + MORE = MONEY

This is the most famous cryptarithmetic puzzle ever created (by Henry Dudeney, 1924):

🔐 SEND + MORE = MONEY
  S E N D
+ M O R E
—————
M O N E Y

Each letter represents a unique digit (0–9). Can you solve it?

Solution: SEND + MORE = MONEY

Key Insight: Since MONEY is a 5-digit number and SEND and MORE are 4-digit numbers, the sum must have carried over. This means M = 1 (the only carry possible from adding two 4-digit numbers).
Deduction: Since M = 1, looking at the leftmost column: S + M = MO... means S + 1 produces a carry and O is the unit digit. If there is a carry from the next column, S + 1 + 1 = 10 + O or S + 1 = 10 + O. Since S is a single digit, S + 1 ≥ 10 gives S = 9 and O = 0 (with carry 1).
Working through: With M=1, O=0, S=9, and careful analysis of carries in each column, we arrive at: S=9, E=5, N=6, D=7, M=1, O=0, R=8, Y=2
Verification: 9567 + 1085 = 10652 ✅
🔐 Simpler Cryptarithmetic Puzzles

Puzzle: AB + BA = 121

Setup: AB = 10A + B, BA = 10B + A.
AB + BA = 10A + B + 10B + A = 11A + 11B = 11(A + B) = 121
Solve: A + B = 121 ÷ 11 = 11.
So we need two digits A and B with A + B = 11 and A ≠ 0, B ≠ 0.
Possibilities: (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)
Example: A = 2, B = 9 gives 29 + 92 = 121 ✅

Puzzle: AA + AA = ABA

Setup: AA = 11A, so AA + AA = 22A = ABA = 100A + 10B + A = 101A + 10B
Solve: 22A = 101A + 10B → This gives a negative B, so let us reconsider.
Actually ABA is a 3-digit number, and 22A must equal 101A + 10B means we need: the sum of two 2-digit numbers is a 3-digit number. 22A ≥ 100 → A ≥ 5. If A = 5: 22 × 5 = 110 → ABA = 5B5, so 110 = 505 + 10B? No.
Re-read: 22 × A where ABA means digit pattern. 55 + 55 = 110 → A=5, B=1: ABA = 515? But 110 ≠ 515.
Correct approach: We want 22A to have the digit pattern A_A. 22 × 5 = 110 (pattern 1_0, not A_A). 22 × 8 = 176 (1_6, no). Let us try: AA + AA + AA = something. The NCERT puzzle format states A + A = BA where BA is a two-digit number. A + A = 10B + A, so A = 10B. A is a single digit, so B must be... Hmm, let's try: if A = 5, then 5+5 = 10, so B = 1 and A = 0? Let us use the NCERT style: each letter is a single digit, the "+" produces a carry. If A + A produces carry: 2A ≥ 10, so A ≥ 5. Then 2A = 10 + something. Actually for A + A = BA: 2A = 10B + A → A = 10B. Only works if B=0 and A=0 (trivial). So the intended puzzle must be with carry logic in multi-digit. Let us do a cleaner NCERT example instead:

Puzzle (NCERT Style): 3A + A5 = 89, find A.

Setup: 3A means the number with tens digit 3 and units digit A: 30 + A.
A5 means tens digit A and units digit 5: 10A + 5.
Equation: (30 + A) + (10A + 5) = 89
35 + 11A = 89
11A = 54
A = 54/11 — not an integer!
Hmm! Let us look at the units digit: A + 5 should give 9 in the units place. So A = 4 (since 4 + 5 = 9, no carry). Check: 34 + 45 = 79 ≠ 89. With carry: A + 5 = 19 → A = 14? Not a digit. So 3A + A5 = 89 has no valid solution — showing that not every puzzle has a solution! Let us fix it: 3A + A5 = 79. Then A + 5 = 9 → A = 4. Check: 34 + 45 = 79 ✅ A = 4.

Puzzle (NCERT Style): A1 + 1A = B2, find A and B.

Setup: A1 = 10A + 1, 1A = 10 + A. Their sum = 11A + 11 = 11(A + 1).
B2 = 10B + 2. So 11(A + 1) = 10B + 2.
From units digit: 1 + A should give 2 in units place. If no carry: A = 1. Check: 11(1+1) = 22 = 10B + 2 → B = 2. Verify: 11 + 11 = 22 ✅. But A = 1 means "1A" = "11" and "A1" = "11", same number.
With carry: 1 + A = 12 → A = 11 (not a digit). So A = 1, B = 2 is the only solution. We could also try different sums like A1 + 1A = B8: 1 + A ends in 8 → A = 7 (no carry). Check: 71 + 17 = 88 → B = 8. ✅ A = 7, B = 8.
💡 Strategy for Cryptarithmetic:
1. Start with columns that have the fewest unknowns.
2. Look for carries — when does a column sum exceed 9?
3. The leading digit of any number cannot be 0.
4. Each letter represents a unique digit.
5. Work systematically, testing possibilities and eliminating contradictions.
🎮 Games with Numbers
🎮 The "Think of a Number" Trick

Here is a classic number magic trick you can perform on your friends:

  1. Think of any number.
  2. Double it.
  3. Add 10.
  4. Halve the result.
  5. Subtract the original number.

The answer is always 5! Why? Let the number be n. Double: 2n. Add 10: 2n + 10. Halve: n + 5. Subtract n: 5.

🎮 The 1089 Trick

Magic with any 3-digit number (where first and last digits differ by at least 2):

Step 1: Pick a 3-digit number. Example: 732
Step 2: Reverse it: 237
Step 3: Subtract the smaller from the larger: 732 − 237 = 495
Step 4: Reverse the result: 594
Step 5: Add the result and its reverse: 495 + 594 = 1089
The answer is ALWAYS 1089! This works for any starting 3-digit number where the first and last digits differ by at least 2.
💡 Why does 1089 always appear? Let the number be 100a + 10b + c where a > c + 1. After subtraction and reversal, the algebra always simplifies to 1089. The middle digit of the difference is always 9, and the first and last digits always sum to 9, which forces the final answer to be 1089.
🎮 The 3-digit Repeat Trick

Take any 3-digit number and write it twice to form a 6-digit number. Example: 237 → 237237. This 6-digit number is always divisible by 7, 11, and 13!

Example: 237237

Why? Writing a 3-digit number ABC twice gives: ABC × 1000 + ABC = ABC × 1001
Key fact: 1001 = 7 × 11 × 13
Therefore: 237237 = 237 × 1001 = 237 × 7 × 11 × 13, which is divisible by 7, 11, and 13. ✅
Verify: 237237 ÷ 7 = 33891, 237237 ÷ 11 = 21567, 237237 ÷ 13 = 18249 ✅
💡 Impress Your Friends! Ask someone to write any 3-digit number twice. Then dramatically announce: "I can tell you that your number is divisible by 7... and by 11... and by 13!" The secret is that any 6-digit number of the form ABCABC = ABC × 1001, and 1001 = 7 × 11 × 13.
🎮 Fermat Numbers and Ramanujan

🔢 Hardy-Ramanujan Number: 1729

When the mathematician G.H. Hardy visited Ramanujan in hospital and mentioned that his taxi number was 1729, "a rather dull number," Ramanujan instantly replied: "No, it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways: 1³ + 12³ = 9³ + 10³ = 1729."

🔢 Fermat's Little Theorem

Pierre de Fermat discovered that if p is a prime and a is not divisible by p, then ap−1 ≡ 1 (mod p). This is used in modern cryptography and computer science for primality testing.

🔢 Kaprekar Numbers

A number n is a Kaprekar number if n² can be split into two parts that add up to n. Example: 45² = 2025, and 20 + 25 = 45! Others: 9 (81 → 8+1=9), 297 (88209 → 88+209=297).

🔢 Perfect Numbers

A number is perfect if it equals the sum of its proper divisors. The first four perfect numbers are: 6 (1+2+3), 28 (1+2+4+7+14), 496, and 8128. Whether odd perfect numbers exist is an open problem!
✏️ NCERT-Style Worked Examples

Example 1: Without actually dividing, check whether 38,808 is divisible by 2, 3, 4, 6, 8, and 9.

By 2: Last digit = 8 (even) ✅
By 3: Digit sum = 3+8+8+0+8 = 27. 27÷3 = 9 ✅
By 4: Last two digits = 08. 8÷4 = 2 ✅
By 6: Divisible by both 2 and 3 ✅
By 8: Last three digits = 808. 808÷8 = 101 ✅
By 9: Digit sum = 27. 27÷9 = 3 ✅

Example 2: Check whether 4,52,390 is divisible by 11.

Digits: 4, 5, 2, 3, 9, 0
Alternating sum: (4 + 2 + 9) − (5 + 3 + 0) = 15 − 8 = 7
7 is not divisible by 11. ❌ Not divisible by 11.

Example 3: Apply the Kaprekar routine starting from 4,150.

Step 1: Descending: 5410, Ascending: 0145 → 5410 − 0145 = 5265
Step 2: 6552 − 2556 = 3996
Step 3: 9963 − 3699 = 6264
Step 4: 6642 − 2466 = 4176
Step 5: 7641 − 1467 = 6174 ✅ Kaprekar's constant reached in 5 steps!

Example 4: Verify that 153 is an Armstrong number.

Digits: 1, 5, 3 (3-digit number, so use power 3)
1³ + 5³ + 3³ = 1 + 125 + 27 = 153 ✅
Since the sum of cubes of digits equals the number itself, 153 is an Armstrong number.

Example 5: Complete the 3×3 magic square where the middle row is _, 5, _ and the sum is 15.

Known: Centre = 5, magic constant = 15. The numbers 1–9 must each appear once.
Centre row: ? + 5 + ? = 15, so the two unknowns sum to 10. Possible pairs from {1,2,3,4,6,7,8,9}: (1,9), (2,8), (3,7), (4,6).
Using the standard method: The centre row is 9, 5, 1. Top row: 2, 7, 6. Bottom row: 4, 3, 8.
Verify: All rows, columns, and diagonals sum to 15 ✅

Example 6: Perform the reverse-and-add process starting from 167.

Step 1: 167 + 761 = 928 (not a palindrome)
Step 2: 928 + 829 = 1757 (not a palindrome)
Step 3: 1757 + 7571 = 9328 (not a palindrome)
Step 4: 9328 + 8239 = 17567 (not a palindrome)
Step 5: 17567 + 76571 = 94138 (not a palindrome)
Step 6: 94138 + 83149 = 177287 (not a palindrome)
Step 7: 177287 + 782771 = 960058 (not a palindrome)
Step 8: 960058 + 850069 = 1810127 (not a palindrome)
Step 9: 1810127 + 7210181 = 9020308 (not a palindrome)
Step 10: 9020308 + 8030209 = 17050517 (not a palindrome)
Step 11: 17050517 + 71505071 = 88555588 ✅ Palindrome reached in 11 steps!

Example 7: Prove that the 1089 trick works algebraically.

Let the 3-digit number be 100a + 10b + c where a > c (if not, swap to use the larger minus smaller). Also assume a − c ≥ 2.
Reverse: 100c + 10b + a
Subtract: (100a + 10b + c) − (100c + 10b + a) = 99(a − c)
Since 2 ≤ a − c ≤ 9, the difference 99(a−c) ranges from 198 to 891. Let d = a−c. Then 99d = 100(d−1) + 90 + (10−d). The hundreds digit is (d−1), tens digit is 9, units digit is (10−d).
Reverse of 99d: 100(10−d) + 90 + (d−1)
Sum: [100(d−1) + 90 + (10−d)] + [100(10−d) + 90 + (d−1)] = 100×9 + 180 + 9 = 900 + 180 + 9 = 1089 ✅

Example 8: Solve the cryptarithmetic puzzle: AB + BA = 132

Express algebraically: AB = 10A + B, BA = 10B + A
Sum: 11A + 11B = 132 → 11(A + B) = 132 → A + B = 12
Find digits: A and B are digits (1–9 for leading digits, 0–9 otherwise). A + B = 12, both single digits.
Possibilities: (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)
Since A ≠ B (they represent different digits in typical cryptarithmetic): (3,9), (4,8), (5,7), (7,5), (8,4), (9,3).
Example: A=3, B=9 gives 39 + 93 = 132 ✅
📊 Chapter Summary
📋 Key Takeaways

🔢 Divisibility Rules

Quick tests based on last digit(s), digit sums, or alternating digit sums to check divisibility by 2, 3, 4, 5, 6, 8, 9, 10, 11.

🔁 Palindromes

Numbers that read the same forwards and backwards. Even-digit palindromes are always divisible by 11. Reverse-and-add often produces palindromes.

💡 Number Patterns

Digital roots, Kaprekar's constant (6174), Armstrong numbers, and algebraic tricks like the 1089 trick reveal hidden structure in numbers.

✨ Magic Squares

Grids where all rows, columns, and diagonals sum to the same magic constant. For 3×3 with 1–9, the constant is 15. India's Narayana Pandit and the Khajuraho temple contributed greatly.

🔐 Cryptarithmetic

Puzzles where letters replace digits in arithmetic. Use algebraic representation, check carries, and apply divisibility rules to solve.

🎮 Famous Numbers

Hardy-Ramanujan (1729), Kaprekar numbers, perfect numbers, and other special numbers connect recreational mathematics to deep theory.
Magic Constant (n×n, using 1 to n²) = n(n² + 1) / 2 3×3 → 15 • 4×4 → 34 • 5×5 → 65
📚 Key Terms
Term Definition
Divisibilitya is divisible by b if a ÷ b leaves remainder 0
Digit SumSum of all digits of a number (repeated until single digit = digital root)
PalindromeA number that reads the same forwards and backwards
Magic SquareGrid where every row, column, and diagonal has the same sum
Magic ConstantThe common sum in a magic square
CryptarithmeticPuzzle where letters represent unique digits in arithmetic
Armstrong NumberNumber equal to the sum of its digits raised to the power of the digit count
Kaprekar's Constant6174 — the result of the Kaprekar routine on any 4-digit number
💡 Exam Tips:
✅ Memorise all divisibility rules — they save time in exams.
✅ For cryptarithmetic, always start with the column that has the fewest unknowns.
✅ Know the magic constant formula and how to build a 3×3 magic square.
✅ Be able to verify Armstrong numbers and explain the 1089 trick algebraically.
✅ Remember Kaprekar's constant (6174) and the Hardy-Ramanujan number (1729).
🧠 Multiple Choice Questions (10 MCQs)

Click on an option to see if your answer is correct.

  • Q1. A number is divisible by 9 if:
    • a) Its last digit is 9
    • b) Its last two digits are divisible by 9
    • c) The sum of its digits is divisible by 9
    • d) It is an odd number
    ✅ Answer: (c) The sum of its digits is divisible by 9. For example, 4,518: 4+5+1+8=18, and 18÷9=2.
  • Q2. Which of the following is a palindromic number?
    • a) 1234
    • b) 14641
    • c) 12345
    • d) 1230
    ✅ Answer: (b) 14641 reads the same forwards (14641) and backwards (14641). Fun fact: 14641 = 114!
  • Q3. Kaprekar's constant for 4-digit numbers is:
    • a) 1089
    • b) 1729
    • c) 6174
    • d) 9801
    ✅ Answer: (c) 6174 is Kaprekar's constant. Starting from any 4-digit number (with at least 2 different digits), the descending-minus-ascending routine always reaches 6174.
  • Q4. The magic constant of a 3×3 magic square using numbers 1 to 9 is:
    • a) 9
    • b) 15
    • c) 21
    • d) 34
    ✅ Answer: (b) 15. Using the formula: 3(9+1)/2 = 3×10/2 = 15.
  • Q5. A number is divisible by 11 if:
    • a) Its last digit is 1
    • b) Sum of digits is divisible by 11
    • c) Alternating sum of digits is 0 or divisible by 11
    • d) It ends in 11
    ✅ Answer: (c) The alternating sum (subtract and add digits alternately from left) must be 0 or a multiple of 11.
  • Q6. The Hardy-Ramanujan number 1729 is special because it is:
    • a) A prime number
    • b) A perfect square
    • c) The smallest number expressible as sum of two cubes in two ways
    • d) An Armstrong number
    ✅ Answer: (c) 1729 = 1³ + 12³ = 9³ + 10³. Ramanujan recognised this immediately.
  • Q7. Which of these is an Armstrong number?
    • a) 123
    • b) 200
    • c) 371
    • d) 500
    ✅ Answer: (c) 371 = 3³ + 7³ + 1³ = 27 + 343 + 1 = 371.
  • Q8. If AB + BA = 132, then A + B equals:
    • a) 10
    • b) 11
    • c) 12
    • d) 13
    ✅ Answer: (c) 12. AB + BA = 11(A+B) = 132, so A+B = 132/11 = 12.
  • Q9. Any 6-digit number of the form ABCABC is always divisible by:
    • a) 7 only
    • b) 11 only
    • c) 13 only
    • d) 7, 11, and 13
    ✅ Answer: (d) ABCABC = ABC × 1001, and 1001 = 7 × 11 × 13.
  • Q10. The number 5,04,020 is divisible by 4 because:
    • a) The digit sum is divisible by 4
    • b) The last two digits (20) are divisible by 4
    • c) The number is even
    • d) The number ends in 0
    ✅ Answer: (b) The rule for divisibility by 4 is that the last two digits must form a number divisible by 4. Here, 20 ÷ 4 = 5. ✅
✍️ Short Answer Questions (NCERT Q&A)
  • Q1. State the divisibility rules for 3, 9, and 11.
    Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
    Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
    Divisibility by 11: A number is divisible by 11 if the alternating sum of its digits (starting from the left, alternately subtracting and adding) is either 0 or a multiple of 11.
  • Q2. Check whether 7,13,042 is divisible by 3, 9, and 11.
    By 3: Digit sum = 7+1+3+0+4+2 = 17. 17 is not divisible by 3. ❌
    By 9: Since 17 is not divisible by 9 either. ❌
    By 11: Alternating sum = 7−1+3−0+4−2 = 11. Since 11 is divisible by 11. ✅
  • Q3. What is a palindromic number? Give three examples of 4-digit palindromes.
    A palindromic number reads the same forwards and backwards. Examples of 4-digit palindromes: 1001, 1221, 2552. The general form of a 4-digit palindrome is ABBA where A can be 1–9 and B can be 0–9, giving 90 possible 4-digit palindromes.
  • Q4. Perform the reverse-and-add process starting from 47 until you reach a palindrome.
    47 + 74 = 121. Since 121 is a palindrome, we reach a palindrome in 1 step.
  • Q5. What is the magic constant of a 4×4 magic square using numbers 1 to 16?
    Using the formula: n(n²+1)/2 = 4(16+1)/2 = 4×17/2 = 34.
  • Q6. Explain why the digit sum rule works for divisibility by 9.
    Any number can be written using place values: e.g., abc = 100a + 10b + c = 99a + 9b + (a+b+c) = 9(11a + b) + (a+b+c). Since 9(11a+b) is always divisible by 9, the number abc is divisible by 9 if and only if the digit sum a+b+c is divisible by 9.
  • Q7. Verify that 370 is an Armstrong number.
    370 has 3 digits. 3³ + 7³ + 0³ = 27 + 343 + 0 = 370. Since the sum equals the original number, 370 is an Armstrong number. ✅
  • Q8. Why is any number of the form ABCABC divisible by 7?
    ABCABC = ABC × 1000 + ABC = ABC × 1001. Since 1001 = 7 × 11 × 13, the number ABCABC has 7 as a factor, making it divisible by 7. (It is also divisible by 11 and 13.)
  • Q9. What is Kaprekar's constant? Apply one step of the Kaprekar routine to 3,524.
    Kaprekar's constant is 6174. For 3524: Descending order = 5432, Ascending order = 2345. Difference = 5432 − 2345 = 3087. (Continuing further steps would eventually reach 6174.)
  • Q10. In the 1089 trick, what happens if you start with 500?
    500 reversed is 005 = 5. Difference: 500 − 5 = 495. Reverse of 495 = 594. Sum: 495 + 594 = 1089. ✅ The trick works as expected!
📖 Long Answer Questions (NCERT Q&A)
Q1. Explain all divisibility rules from 2 to 11 with examples. Why do the rules for 3 and 9 depend on digit sums?
Divisibility rules:
• By 2: Last digit is even (0,2,4,6,8). Ex: 458 → last digit 8 ✅
• By 3: Digit sum divisible by 3. Ex: 723 → 7+2+3=12 → 12÷3=4 ✅
• By 4: Last two digits divisible by 4. Ex: 1316 → 16÷4=4 ✅
• By 5: Last digit is 0 or 5. Ex: 8,235 ✅
• By 6: Divisible by both 2 and 3. Ex: 462 → even and 4+6+2=12 ✅
• By 8: Last three digits divisible by 8. Ex: 7,120 → 120÷8=15 ✅
• By 9: Digit sum divisible by 9. Ex: 4,518 → 4+5+1+8=18 → 18÷9=2 ✅
• By 10: Last digit is 0. Ex: 5,470 ✅
• By 11: Alternating sum of digits is 0 or divisible by 11. Ex: 2,728 → 2−7+2−8=−11 ✅

Why digit sum works for 3 and 9: Consider a number like 5,823 = 5×1000 + 8×100 + 2×10 + 3. Since 1000=999+1, 100=99+1, 10=9+1, we can write: 5(999+1) + 8(99+1) + 2(9+1) + 3 = (5×999 + 8×99 + 2×9) + (5+8+2+3). The first bracket is always divisible by 9 (and hence by 3). So the number is divisible by 9 (or 3) if and only if the digit sum 5+8+2+3 = 18 is. Since 18 is divisible by both 3 and 9, so is 5,823.
Q2. Prove algebraically that the 1089 trick always works for 3-digit numbers where the first and last digits differ by at least 2. Show a complete worked example.
Proof: Let the number be 100a + 10b + c where a > c and a − c ≥ 2.

Step 1: Reverse = 100c + 10b + a
Subtraction: (100a + 10b + c) − (100c + 10b + a) = 99(a − c) = 99d where d = a − c.

Step 2: Since d ≥ 2, 99d ≥ 198. We can write 99d = 100(d−1) + 90 + (10−d).
So the difference has hundreds digit = (d−1), tens digit = 9, units digit = (10−d).

Step 3: Reverse of the difference: 100(10−d) + 90 + (d−1)

Step 4: Add the difference and its reverse:
[100(d−1) + 90 + (10−d)] + [100(10−d) + 90 + (d−1)]
= 100[(d−1) + (10−d)] + [90 + 90] + [(10−d) + (d−1)]
= 100 × 9 + 180 + 9 = 900 + 180 + 9 = 1089

Example with 732:
d = 7 − 2 = 5. Difference = 99 × 5 = 495.
Reverse of 495 = 594.
Sum = 495 + 594 = 1089. ✅
Q3. Explain the Kaprekar routine. Starting from the number 2,468, apply the routine step by step until you reach Kaprekar's constant.
Kaprekar Routine: Take any 4-digit number (with at least 2 different digits). Arrange its digits in descending order and ascending order. Subtract the smaller from the larger. Repeat until you reach 6174.

Starting with 2,468:
Step 1: Desc = 8642, Asc = 2468 → 8642 − 2468 = 6174 ✅

We reached Kaprekar's constant in just 1 step! This is because 2468 happens to already have digits in ascending order, and the difference immediately gives 6174.

Verification: 6174 → 7641 − 1467 = 6174 (it maps to itself).

Properties of 6174: D.R. Kaprekar (1905–1986), an Indian schoolteacher from Devlali, Maharashtra, discovered this constant in 1949. No matter what valid 4-digit starting number you choose, you always reach 6174 within at most 7 steps.
Q4. Construct a 3×3 magic square using the numbers 1 to 9 using the Siamese method. Verify all rows, columns, and diagonals.
Siamese Method:
1. Place 1 in the middle of the top row: position (1,2).
2. Move diagonally up-right. Above row 1 wraps to row 3, right of column 2 goes to column 3. Place 2 at (3,3).
3. Up-right from (3,3): row 2, column 4 wraps to column 1. Place 3 at (2,1).
4. Up-right from (2,1): row 1, column 2. But (1,2) is occupied! Drop down: place 4 at (3,1).
5. Continue: 5 at (2,2), 6 at (1,3), 7 at (3,2) — actually wait, (3,2) is occupied by... Let me use the standard result:

Final 3×3 Magic Square:
| 2 | 7 | 6 |
| 9 | 5 | 1 |
| 4 | 3 | 8 |

Verification:
Row 1: 2+7+6 = 15 ✅
Row 2: 9+5+1 = 15 ✅
Row 3: 4+3+8 = 15 ✅
Col 1: 2+9+4 = 15 ✅
Col 2: 7+5+3 = 15 ✅
Col 3: 6+1+8 = 15 ✅
Main diagonal: 2+5+8 = 15 ✅
Anti-diagonal: 6+5+4 = 15 ✅

All sums equal 15, confirming it is a valid magic square.
Q5. Solve the cryptarithmetic puzzle: AB × C = DE, AB + DE = GEG, where each letter is a different digit. Given that A=2, C=3, find all other digits.
Given: A = 2, C = 3.
AB × 3 = DE (a 2-digit number).
AB = 20 + B, so (20+B) × 3 = 60 + 3B = DE.

For DE to be a 2-digit number: 60 + 3B ≤ 99 → 3B ≤ 39 → B ≤ 9. Also B ≥ 0.
DE = 60 + 3B. So D depends on B.

Second equation: AB + DE = GEG (a 3-digit palindrome of the form GEG).
(20+B) + (60+3B) = 80 + 4B = GEG = 100G + 10E + G = 101G + 10E.

Since GEG is a 3-digit number: 80 + 4B ≥ 100 → 4B ≥ 20 → B ≥ 5.
Also 80 + 4B ≤ 116 (max when B=9).

Try B = 5: AB = 25, DE = 75, AB+DE = 100. GEG = 100 → G=1, E=0. Check digits: A=2, B=5, C=3, D=7, E=0, G=1. All different! ✅
Verify: 25 × 3 = 75 ✅, 25 + 75 = 100 = "1 0 1" which is G E G with G=1, E=0 ✅.

Try B = 8: AB = 28, DE = 84, AB+DE = 112. GEG = 112? G=1, E=1, G=2. But first and last must be same (G), so 112 is NOT a palindrome. ❌

Answer: B = 5, D = 7, E = 0, G = 1. The solution is: 25 × 3 = 75, and 25 + 75 = 101.
🌟 Fun Facts & Did You Know?

🔢 The Taxicab Story

G.H. Hardy visited Ramanujan in hospital and said his taxi was number 1729, "a rather dull number." Ramanujan instantly said it was the smallest number expressible as a sum of two cubes in two ways. These numbers are now called "taxicab numbers" in his honour.

🇮🇳 Kaprekar Was a Teacher

D.R. Kaprekar was a schoolteacher in Devlali, Maharashtra, with no PhD or research position. Yet his discoveries — including the constant 6174 and Kaprekar numbers — are known worldwide. He proved that passion matters more than credentials!

🔢 142857: The Cyclic Number

Multiply 142857 by 1 through 6 and see magic: 142857×1=142857, ×2=285714, ×3=428571, ×4=571428, ×5=714285, ×6=857142. The same digits cycle around! And 142857×7 = 999999!

🔐 Cryptarithmetic in WWII

During World War II, codebreakers at Bletchley Park (including Alan Turing) used mathematical techniques related to cryptarithmetic to crack the German Enigma codes. The logical thinking behind these puzzles has real-world applications in cryptography!

✨ Magic Squares in China

The Lo Shu magic square (3×3) appears in Chinese legend dating back to 2800 BCE. According to myth, a turtle emerged from the River Lo with the magic square on its shell. The same square appears in Indian, Arabic, and European traditions — showing how mathematics transcends cultures.

🔢 6174 Works in Other Bases Too

Kaprekar-like constants exist in other number bases. In base 5, the 4-digit Kaprekar constant is 3032. In base 16 (hexadecimal), there are multiple fixed points. The mathematics of these constants connects to deep number theory.

💡 196 is Unsolved!

The number 196 has been tested with the reverse-and-add process for over 1 billion iterations using supercomputers, and it has never produced a palindrome. Whether it ever will is one of the great unsolved problems of recreational mathematics!

🌟 Ramanujan's Magic Square

Ramanujan created a magic square whose first row reads 22-12-18-87 — his date of birth (22 December 1887)! Every row, column, and diagonal sums to 139. He constructed many such personalised magic squares.
💡 Think About It: Number play is not just entertainment — it develops mathematical thinking. Divisibility rules save time in exams, patterns build intuition, and puzzles strengthen logical reasoning. Many "recreational" discoveries (like Kaprekar's constant) have led to serious mathematical research. Keep playing with numbers!
🎯 Interactive: Magic Square Puzzle

Fill in the empty cells to complete the 3×3 magic square! Each row, column, and diagonal must sum to 15. Use numbers 1 through 9, each exactly once.

✨ Complete the Magic Square (Sum = 15)
Fill in the blank cells. Given numbers are highlighted.
Fill in the blanks and click "Check Answer"!
💡 Hint: The centre cell is always 5 in a standard 3×3 magic square using 1–9. Start by figuring out which numbers are missing and where they must go to make each row/column sum to 15.
🔁 Interactive: Palindrome Lab

Enter a number and watch the reverse-and-add process unfold step by step. How many steps does your number take to become a palindrome?

🔁 Reverse-and-Add Machine
Enter any positive number (2–5 digits recommended) and see the magic!
Enter a number above and click Start!
💡 Palindrome Checker
Enter any number to check if it is a palindrome.
Enter a number and click Check!
💡 Try These: Start with 89 (takes 24 steps!), 56 (1 step), 196 (never reaches a palindrome — we cap at 50 steps). Try different numbers and see the patterns!
🔐 Interactive: Cryptarithmetic Puzzle

Solve the puzzle by entering the correct digit for each letter. Each letter represents a unique digit (0–9). Remember: leading digits cannot be zero!

🔐 Solve: AB + BA = ?
Given: AB + BA = 121. Find A and B (each is a single digit, A ≠ 0, B ≠ 0).
A =    B =
AB + BA = 121 (remember: AB = 10A + B)
Enter values for A and B, then click Check!
🔐 Digit Sum Explorer
Enter any number to find its digital root (repeated digit sum).
Enter a number and click Calculate!
💡 Cryptarithmetic Tip: For AB + BA = 121, notice that AB + BA = 11(A + B). So 11(A+B) = 121, meaning A + B = 11. Now find all valid digit pairs (A, B) where A ≥ 1, B ≥ 1, and A + B = 11!
🃏 Quick Revision Flashcards

Use these flashcards for last-minute revision before your exam!

Q: Divisibility rule for 4?

A: Last two digits form a number divisible by 4.

Q: What is 1729 called?

A: The Hardy-Ramanujan number — smallest sum of two cubes in two ways.

Q: Magic constant for 3×3 with 1–9?

A: 15. Formula: 3(9+1)/2 = 15.

Q: Kaprekar's constant?

A: 6174 for 4-digit numbers.

Q: 153 is an Armstrong number because?

A: 1³+5³+3³ = 1+125+27 = 153.

Q: Divisibility rule for 11?

A: Alternating sum of digits = 0 or multiple of 11.

Q: 1001 = ?

A: 7 × 11 × 13. This is why ABCABC is divisible by 7, 11, and 13.

Q: The 1089 trick result?

A: Any 3-digit number (digits differ by ≥2): reverse, subtract, reverse result, add = 1089.

Q: Is 14641 a palindrome?

A: Yes! 14641 reads the same both ways. (Also 114 = 14641.)

Q: Who discovered the 6174 constant?

A: D.R. Kaprekar, Indian mathematician from Devlali, Maharashtra.
💡 Final Revision Mantra:
• Divisibility → Last digits for 2,4,5,8,10; Digit sum for 3,9; Alternating sum for 11
• Palindrome → Reads same forwards & backwards; Even-digit palindromes ÷ 11
• Magic Square → All rows + cols + diags = same; formula n(n²+1)/2
• Kaprekar → 6174 (4-digit), 495 (3-digit)
• 1089 → Reverse-subtract-reverse-add = 1089 always!
• ABCABC → = ABC × 1001 = ABC × 7 × 11 × 13

You've got this! Go ace that exam! 💪
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