Mathematics Class 6 — Ganita…
🏠 Home Log in
TOPIC 10 OF 41

Kaprekar’s Constant, Clock Arithmetic, Mental Math and Collatz

🎓 Class 6 Mathematics CBSE Theory Ch 3 — Number Play ⏱ ~35 min
🌐 Language: [gtranslate]

This MCQ module is based on: Kaprekar’s Constant, Clock Arithmetic, Mental Math and Collatz

This mathematics assessment will be based on: Kaprekar’s Constant, Clock Arithmetic, Mental Math and Collatz
Targeting Class 6 level in Number Theory, with Basic difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

3.6 The Magic Number of Kaprekar

D.R. Kaprekar? was a mathematics teacher in Devlali, Maharashtra. In 1949, he discovered a fascinating pattern in 4-digit numbers.

Kaprekar's Process
Pick any 4-digit number having at least two different digits (e.g. 6382). Then:
  1. Arrange its digits in descending order → call it A.
  2. Arrange its digits in ascending order → call it B.
  3. Compute C = A − B.
  4. Repeat with C.
Within a few steps (usually ≤ 7), you always reach 6174 — Kaprekar's Constant!
Start: 6382 A = 8632(descending) B = 2368(ascending) C = 6264 6642 − 2466 = 4176 7641 − 1467 = 6174 7641 − 1467 = 6174 ✓
Kaprekar chain for 6382: 6264 → 4176 → 6174 (reached in 3 steps).

3.7 Clock and Calendar Numbers

On a 12-hour clock, timings contain digit patterns. For example, 4:44, 10:10, 12:21 look special. Let's hunt for more.

Q (p. 64) — Try and find all possible times on a 12-hour clock of each of these types: Manish has his birthday on 20/12/2012 where the digits '2', '0', '1', and '2' repeat in the date itself. Find some other dates of this form from the past.
Dates using only digits {0, 1, 2}: 01/02/2010, 10/02/2001, 02/10/2012, 21/01/2020, 02/01/2012, 20/01/2012 … Any date dd/mm/yyyy where every digit ∈ {0,1,2}.
His sister, Meghana, has her birthday on 11/02/2011 where the digits read the same from left-to-right and from right-to-left. Find all possible dates of this form from the past.
Palindromic 8-digit dates ddmmyyyy: 10/02/2001, 20/02/2002, 01/02/2010, 11/02/2011, 21/02/2012, 02/02/2020, 12/02/2021, 22/02/2022, 03/02/2030, 13/02/2031, … These all read the same both ways.
Q. Jeevan was looking at this year's calendar. Why should we change the calendar every year? Can we not reuse a calendar?
The day on which any date falls shifts by 1 day each normal year and 2 days in a leap year. So calendars can be reused after 6, 11, 28 years etc. (they repeat in a cycle).
Q. But will any year's calendar repeat again after some years? Will all dates and days in a year match exactly with that of another year?
Yes. A non-leap year calendar repeats every 6 or 11 years. A leap-year calendar repeats every 28 years.

Figure it Out (Section 3.7)

Q1. Pratibha uses digits 4, 7, 3, 2 and makes smallest and largest 4-digit numbers. Difference between these two numbers is 7432 − 2347 = 5085. The sum of digits in both 7432 and 2347 is 4 + 7 + 3 + 2 = 16. Now 5085 has digits sum 5 + 0 + 8 + 5 = 18. (a) Can you find digits which give the largest and smallest sums? (b) Less than 5085?
(a) Largest gap: use 9, 8, 2, 1 → 9821 − 1289 = 8532. Largest gap = 8532 using digits 9, 9, 0, 0 → 9900 − 0099 = 9801.
(b) Smallest difference with 4 distinct digits: 2134 − 1234 = 900 using digits 1,2,3,4 (consecutive). Or using 4, 7, 3, 2 if we pick other digits with closer largest-smallest swap you can find differences less than 5085.

3.8 Mental Math

Look at the middle-column numbers 25,000, 400, 13,000, 1,500, and 60,000 all formed from side numbers like 38,800; 5,400; 28,000; 63,000 etc. The rule: side numbers are sums/differences of neighbours in the middle.

Example
38,800 = 25,000 + 400 × 2 + 13,000
3400 + 1500 = 1500 + 400 = 400
Can we make 1,000 using the numbers in the middle? Why or why not?
Numbers in the middle: 25,000, 400, 13,000, 1500, 60,000. We cannot reach 1000 using positive addition alone of these values since their minimum is 400 and the next is 1500 — skipping over 1000. However 1500 − 400 − some combination won't equal exactly 1000 with only these terms once. So No, with the given middle numbers we cannot make 1000. But we can make 14,000; 15,000 and 16,000: yes. Explore further.
Adding/subtracting: 39,800 = 40,000 − 800 × 300 + 300 = 45,000 − 5900 = 17,500 + 21,400. Verify.
Each expression rearranges the computation into easier sub-problems. Mental math thrives on regrouping: 39,800 = 40,000 − 200 (easy). 17,500 + 21,400 = 38,900 (not 39,800 — the textbook equation illustrates the strategy, not exact equality).

Figure it Out (Section 3.8)

Q1. Write an example for each of the scenarios: 5-digit + 5-digit = 6-digit (more than 90,250); 5-digit + 3-digit = 6-digit; 4-digit + 4-digit = 6-digit; 5-digit − 5-digit = 1-digit (gives 18,500); 5-digit − 4-digit = 5-digit; 5-digit − 5-digit = 4-digit (less than 56,503); 4-digit × 4-digit = 6-digit; 5-digit − 3-digit = 4-digit (gives 91,500); 5-digit − 3-digit = 3-digit.
Sample:
• 92,000 + 95,000 = 187,000 (5+5→6 digit)
• 99,500 + 800 = 1,00,300 (5+3→6)
• 9500 + 9500 = 19,000 (4+4 can't reach 6-digit → impossible). The scenario "4-digit + 4-digit = 6-digit" is not possible (max 9999+9999 = 19,998 which is 5-digit).
• 50,000 − 49,999 = 1 (1-digit)
• 50,000 − 1000 = 49,000 (5−4→5)
• 56,000 − 55,000 = 1,000 vs asked <56,503: choose 56,500 − 56,400 = 100 (<1000) gives 3-digit; for 4-digit use 56,500 − 54,000 = 2,500 ✓
• 1500 × 700 = 10,50,000 — actually 4-digit × 4-digit = up to 8-digit; one 6-digit example: 1234 × 100 = no (100 is 3-digit). 3456 × 100 not allowed. 1000 × 1000 = 10,00,000 (7-digit). Try 3456 × 30 — not 4×4. Typically 4×4 gives at least 7 digits.
• 91,500 + 200 = 91,700 (reverse: 91,700 − 200 = 91,500) ✓
• 10,500 − 500 = 10,000 → 3-digit result impossible since subtraction of a 3-digit from 5-digit always gives at least 5-digit or 4-digit result… (some cases impossible — demonstrates number-size logic).

3.9 Playing with Number Patterns

Arrangements like:

   40   40   40   40
 50   50   50   50   50
   40   40   40   40
 50   50   50   50   50
   40   40   40   40
Pattern (a) — sum = 4×40×3 + 5×50×2? Clever grouping helps.

Method A (pair rows): Each pair (50-row + 40-row) = 5×50 + 4×40 = 250 + 160 = 410. With 2 such pairs plus extra 40-row → 2×410 + 160 = 820 + 160 = 980.

3.10 An Unsolved Mystery — The Collatz Conjecture!

History
In the 1930s, the German mathematician Lothar Collatz? conjectured that the following sequence will always reach 1, regardless of the whole number you start with:
  • If the number is even, take half of it.
  • If the number is odd, multiply by 3 and add 1.
  • Repeat.
Even today, mathematicians working on this — it remains an unsolved problem in mathematics. Collatz's conjecture is true of the most famous unsolved problems in mathematics.

Examples of Collatz sequences:

  • 12 → 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1
  • 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
  • 21 → 64 → 32 → 16 → 8 → 4 → 2 → 1
  • 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

Figure it Out

Make more Collatz sequences starting with your favourite whole numbers. Do you always reach 1? Why or why not?
Yes — for every starting number tested so far, the sequence reaches 1. But no general proof exists. It is an open problem. All numbers up to 268 have been verified by computers.

3.11 Simple Estimation

At times, we may not know or need an exact count of things, and an estimate is sufficient for the purpose at hand. A class teacher might know roughly 500 students are enrolled in the school, but may only know an estimated count.

Paromita's class has 32 children. Other sections are similar. So she estimated the number of children in her class to be 32 × 10 = 320. Class 5 has 29 students, and each class has 3 sections of about 500. Estimate the total number of children in the school: if about 29 students in Class 6, 7, 8 etc. Is 500 reasonable?
If each class has 3 sections × ~30 students = 90 students per class. From classes 1 to 10 → ~900 students. So estimate 500 may be low — depends on the actual school.
(Figure it Out — Estimate the answer guessing in 30 seconds) 1. Number of words in your maths textbook: more than 5000 / less than 5000? 2. Number of students in your school travelling by bus: more than 200 / less than 200? 3. Roshan wants to buy milk and fruit to make fruit custard for 5 people. He estimates the cost as ₹100. Do you agree?
1. Most Class 6 textbooks have more than 5000 words.
2. Answer depends — in most schools, less than 200 take buses unless large.
3. Fruit custard for 5 (≈₹1-1.5 litre milk ₹60, fruits ₹40) ≈ ₹100, Yes this is a reasonable estimate.

3.12 Games and Winning Strategies

Numbers can be used to play games and develop winning strategies.

Rules — Game #1 (21)
Starting at 0, two players take turns adding 1, 2 or 3 to the running total. The player who reaches 21 wins.
Winning strategy for Game #1: The key numbers are 1, 5, 9, 13, 17, 21 — a ladder with step 4. If you can say 17, then whatever opponent adds (1, 2, or 3), you adjust to land on 21. So aim for 17, then 13, then 9, etc. The first player always wins by saying 1 first, then mirroring to reach the ladder.
Rules — Game #2 (99)
Two players take turns adding a number between 1 and 10 to the running total. The first player to reach 99 wins.

Strategy: Step is (1+10) = 11. Target ladder: 99, 88, 77, 66, 55, 44, 33, 22, 11. The first player says 11 first, then mirrors. First player wins.

Activity: Predict → Observe → Explain — Kaprekar's Magic
Predict: How many steps does it take for 6382 to reach 6174? Will every 4-digit number reach 6174?
  1. Pick any 4-digit number with at least two different digits.
  2. Form the largest and smallest arrangements of its digits.
  3. Subtract and record the result.
  4. Repeat until you reach 6174. Count the steps.
  5. Try it for 10 different numbers.
Observe: Every 4-digit number (with at least 2 distinct digits) reaches 6174 in at most 7 iterations. Once you reach 6174, 7641 − 1467 = 6174 again — a fixed point. Explain: 6174 has a special property: its own Kaprekar step returns itself. All other paths feed into it.
Competency-Based Questions
Scenario: Meera explores Kaprekar's process and also plays the 21-game with her cousin. She wonders about patterns in both.
Q1. Apply Kaprekar's process to the number 3524. Reach 6174 and count the number of steps.
L3 Apply
3524: A = 5432, B = 2345 → 5432 − 2345 = 3087. Then: 8730 − 0378 = 8352. Then: 8532 − 2358 = 6174. 3 steps to reach 6174.
Q2. Analyse: Why does the Kaprekar process not work on 3333 (or any repdigit)?
L4 Analyse
A repdigit (e.g. 3333) has all digits identical. Largest = Smallest = 3333. Difference = 0, which is not a 4-digit number. So the process requires at least two different digits.
Q3. Evaluate: "In the 21-game, the player who plays second always loses if both play optimally." True or false? Justify.
L5 Evaluate
True. First player says 1 (landing on ladder), then mirrors any move to keep on the 1, 5, 9, 13, 17, 21 ladder. Second player can never land on 17 or 21 against optimal play.
Q4. Create a new game similar to Game #2: two players take turns adding between 1 and 4 to a running total; first to reach 50 wins. Find a winning strategy.
L6 Create
Step = 1 + 4 = 5. Target ladder: 50, 45, 40, 35, 30, 25, 20, 15, 10, 5. First player says 5 first, then matches opponent's move by 5 − opponent's move each turn. First player always wins.
Assertion–Reason Questions
A: 6174 is called Kaprekar's constant because applying Kaprekar's process to it gives 6174 back.
R: 7641 − 1467 = 6174.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (a) — R is the calculation that shows the fixed-point property of 6174.
A: Starting from 7, the Collatz sequence reaches 1 in 16 steps.
R: Collatz's conjecture has been proved true for all starting numbers.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (c) — 7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 (16 steps). But R is false: the Collatz conjecture is NOT proved — only verified for many numbers.
A: In the 99-game (add 1–10), the first player wins by saying 11 first.
R: The key ladder 11, 22, 33, …, 99 has step size equal to (min + max) = 1 + 10 = 11.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (a) — Both true; the step size explains the ladder strategy.

Frequently Asked Questions

What is Kaprekar's constant?

Kaprekar's constant is 6174. Take any 4-digit number with at least two different digits, arrange its digits in descending then ascending order, and subtract. Repeat this process with the result and you always reach 6174 within 7 steps. This magical property is explored in NCERT Class 6 Ganita Prakash Chapter 3.

How does the Kaprekar routine work with an example?

Take 3524. Largest: 5432. Smallest: 2345. Subtract: 5432 - 2345 = 3087. Repeat: 8730 - 0378 = 8352. Then 8532 - 2358 = 6174. From 6174 onwards: 7641 - 1467 = 6174 (fixed). NCERT Class 6 Chapter 3 demonstrates this pattern.

What is the Collatz conjecture in Class 6 Maths?

The Collatz conjecture says: start with any positive whole number. If it's even, divide by 2. If odd, multiply by 3 and add 1. Keep repeating, and you always eventually reach 1. This unsolved problem is introduced in NCERT Class 6 Ganita Prakash Chapter 3.

What is clock arithmetic in Chapter 3?

Clock arithmetic uses the cyclic nature of hours on a clock. Adding 5 hours to 10 o'clock gives 3 o'clock, not 15. This modulo arithmetic is introduced in NCERT Class 6 Ganita Prakash Chapter 3 as a playful way to think about remainders.

Why learn mental math shortcuts in Class 6?

Mental math shortcuts such as adding by rearranging, using nearby round numbers, or splitting digits make calculations faster and reduce errors. NCERT Class 6 Ganita Prakash Chapter 3 introduces these tricks to build number sense and confidence.

Does Kaprekar's process work for all 4-digit numbers?

Kaprekar's process converges to 6174 for all 4-digit numbers except those with all four digits the same (like 1111, 2222) which give 0. Within at most 7 iterations every qualifying number reaches 6174, as discussed in NCERT Class 6 Chapter 3.

Frequently Asked Questions — Number Play

What is Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz in NCERT Class 6 Mathematics?

Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz is a key concept covered in NCERT Class 6 Mathematics, Chapter 3: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz step by step?

To solve problems on Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 6 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 3: Number Play?

The essential formulas of Chapter 3 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz important for the Class 6 board exam?

Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz is part of the NCERT Class 6 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz?

Common mistakes in Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz?

End-of-chapter NCERT exercises for Kaprekar's Constant, Clock Arithmetic, Mental Math and Collatz cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.

Keyword

AI Tutor
Mathematics Class 6 — Ganita Prakash
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Kaprekar’s Constant, Clock Arithmetic, Mental Math and Collatz. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.