This MCQ module is based on: 5.6 Fun with Numbers — Special Numbers Puzzle
TOPIC 19 OF 41
5.6 Fun with Numbers — Special Numbers Puzzle
🎓 Class 6
Mathematics
CBSE
Theory
Ch 5 — Prime Time
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 5.6 Fun with Numbers — Special Numbers Puzzle
Targeting Class 6 level in Number Theory, with Basic difficulty.
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5.6 Fun with Numbers — Special Numbers Puzzle
Guna writes four 2-digit numbers on the board: 9, 16, 25, 43. He challenges his classmates: "Which of these is special, and why?"
Different viewpoints
- Karnavati: "9 is special — it is the only single-digit number."
- Gurupreet: "9 is special — it is the only multiple of 3."
- Murugan: "16 is special — it is the only even number / multiple of 2."
- Gopika: "25 is special — it is the only multiple of 5."
- Tadnyikee: "43 is special — it is the only prime number."
- Radha: "43 is special — it is the only number that is not a perfect square."
🧩 Try This Puzzle
Design your own set of four numbers so that (i) one is special because it is the only prime, (ii) one is special because it is the only perfect square, (iii) one is special because it is the only multiple of 10. Is there always a single "most special" number? Discuss with a friend.
Chapter 5 — Full Exercises
Here we revise the main competencies of Chapter 5 with a mixed question bank drawn from the full NCERT "Figure it Out" sections.
Q1. At what number is "idli-vada" said for the 10th time in a 3-and-5 game?
Common multiples of 3 and 5 are multiples of 15. 10th → 150.
Q2. In a 3 and 5 idli-vada game played up to 100, which is the smallest number for which the child says "idli"?
The first multiple of 3 is 3.
Q3. Are these pairs co-prime? Use prime factorisation:
(a) 30 and 45 (b) 57 and 85 (c) 121 and 1331 (d) 343 and 216
(a) 30 and 45 (b) 57 and 85 (c) 121 and 1331 (d) 343 and 216
(a) 30 = 2·3·5, 45 = 3²·5 → common 3, 5 → No.
(b) 57 = 3·19, 85 = 5·17 → Yes.
(c) 121 = 11², 1331 = 11³ → common 11 → No.
(d) 343 = 7³, 216 = 2³·3³ → Yes.
(b) 57 = 3·19, 85 = 5·17 → Yes.
(c) 121 = 11², 1331 = 11³ → common 11 → No.
(d) 343 = 7³, 216 = 2³·3³ → Yes.
Q4. Is the first number divisible by the second? Use prime factorisation.
(a) 225, 27 (b) 96, 24 (c) 343, 17 (d) 999, 99
(a) 225, 27 (b) 96, 24 (c) 343, 17 (d) 999, 99
(a) 225 = 3²·5², 27 = 3³ — 225 has only 3², misses one 3 → No.
(b) 96 = 2⁵·3, 24 = 2³·3 — all factors of 24 appear in 96 → Yes.
(c) 343 = 7³, 17 prime — 17 not in 343 → No.
(d) 999 = 3³·37, 99 = 3²·11 — 11 not in 999 → No.
(b) 96 = 2⁵·3, 24 = 2³·3 — all factors of 24 appear in 96 → Yes.
(c) 343 = 7³, 17 prime — 17 not in 343 → No.
(d) 999 = 3³·37, 99 = 3²·11 — 11 not in 999 → No.
Q5. Find the prime factorisation of: (a) 120 (b) 300 (c) 540 (d) 2024.
(a) 120 = 2³·3·5 (b) 300 = 2²·3·5² (c) 540 = 2²·3³·5 (d) 2024 = 2³·11·23
Q6. Is 15,470 divisible by 4? Use the last-two-digit rule.
Last two digits = 70; 70/4 = 17 remainder 2 → No.
Q7. Is 8,576,250 divisible by (i) 3 (ii) 9 (iii) 10?
Digit sum = 8+5+7+6+2+5+0 = 33 → div by 3 yes, by 9 no (33 not div by 9). Last digit 0 → div by 10 yes.
Q8. How many years between 2000 and 2099 (inclusive) are leap years?
Multiples of 4 from 2000 to 2096: (2096−2000)/4 + 1 = 25. 2000 is div by 400 so included; no century-exclusion in this range. 25 leap years.
Q9. Find two numbers whose product is 10000 such that neither ends in zero.
\(10000 = 2^4 \cdot 5^4 = 16 \cdot 625\). Pair: 16 and 625.
Q10. Find the smallest number divisible by every integer from 1 to 10.
LCM(1..10) = 2³·3²·5·7 = 2520.
Q11. Show that 6 is a perfect number.
Factors of 6: 1, 2, 3, 6. Sum of factors excluding itself = 1+2+3 = 6. Hence 6 is perfect.
Q12. Sum of digits of 73,800. Is it divisible by 9?
7+3+8+0+0 = 18 → div by 9 yes.
Chapter 5 — Summary
Multiples & Factors
Multiples of \(k\) are numbers like \(k, 2k, 3k,\ldots\). Factors of \(n\) are numbers that divide \(n\) exactly.Common Multiples/Factors
Numbers that are multiples (or factors) of two numbers simultaneously.Primes & Composites
A number > 1 is prime if its only factors are 1 and itself; otherwise composite. 1 is neither.Co-primes
Two numbers whose only common factor is 1. E.g. 8 and 15.Prime Factorisation
Every number > 1 can be written uniquely as a product of primes (Fundamental Theorem).Divisibility Tests
Quick rules for 2, 3, 4, 5, 6, 8, 9, 10 using last digits or digit-sums.Sieve of Eratosthenes
Method to find all primes up to \(N\) by crossing out multiples.Perfect Numbers
\(n\) whose proper-factor sum equals \(n\). First two: 6 and 28.Key Formulas
• A multiple of both \(a\) and \(b\) is a multiple of LCM(\(a,b\)).• A common factor of \(a\) and \(b\) is a factor of HCF(\(a,b\)).
• \(a\) and \(b\) are co-prime \(\iff\) HCF(\(a,b\)) = 1.
• \(m\) is divisible by \(n\) \(\iff\) every prime in the factorisation of \(n\) appears (with same multiplicity) in \(m\).
Activity: Build-Your-Own Factor Tree
L3 Apply
Materials: Chart paper, coloured pens
Predict: Will the prime factorisation of 144 contain more 2s or 3s? Guess!
- Pick any 3-digit number (say 144).
- Split it into two factors at each step (144 = 12 × 12 or 16 × 9 — both are fine).
- Keep splitting non-prime nodes until all leaves are prime.
- Collect the leaves — that is the prime factorisation!
- Check: try a different first-split. You should get the same multiset of primes.
144 = 2⁴·3². Four 2s and two 3s — more 2s.
Competency-Based Questions
Scenario: A bakery packages exactly 60 cupcakes per tray and 84 biscuits per tray. The owner wants to prepare gift boxes so that each box has the same number of cupcakes and the same number of biscuits, with nothing left over.
Q1. What is the largest number of identical gift boxes she can prepare?
L3 ApplyHCF(60, 84). 60 = 2²·3·5, 84 = 2²·3·7. HCF = 2²·3 = 12 boxes. Each has 5 cupcakes and 7 biscuits.
Q2. If instead she wants to prepare a SINGLE large tray where she can lay both cupcakes and biscuits in equal rows, what is the smallest tray-capacity that works?
L4 AnalyseLCM(60, 84) = 2²·3·5·7 = 420 items per tray.
Q3. Judge: "Since HCF(60, 84) = 12 and LCM = 420, the product 12 × 420 equals 60 × 84." Is this coincidence?
L5 EvaluateNot a coincidence. For any two numbers \(a\) and \(b\), HCF × LCM = \(a \times b\). Check: 12 × 420 = 5040 = 60 × 84.
Q4. Design a packaging scheme involving three items (say 24 chocolates, 36 toffees, 60 candies) so that each gift box receives the same number of each item. How many boxes, and what goes in each?
L6 CreateHCF(24, 36, 60): 24 = 2³·3, 36 = 2²·3², 60 = 2²·3·5. HCF = 2²·3 = 12 boxes. Each box: 2 chocolates, 3 toffees, 5 candies.
Assertion–Reason Questions
A: The LCM of two co-prime numbers equals their product.
R: Co-prime numbers have HCF equal to 1.
R: Co-prime numbers have HCF equal to 1.
Answer: (a) — since HCF × LCM = product, if HCF = 1 then LCM = product.
A: 28 is a perfect number.
R: The sum of all factors of 28 (including 28) equals 56.
R: The sum of all factors of 28 (including 28) equals 56.
Answer: (a) — σ(28) = 56 = 2·28, which means proper-divisor sum equals 28.
Frequently Asked Questions — Prime Time
What is Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool in NCERT Class 6 Mathematics?
Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool is a key concept covered in NCERT Class 6 Mathematics, Chapter 5: Prime Time. 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 Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool step by step?
To solve problems on Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool, 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 5: Prime Time?
The essential formulas of Chapter 5 (Prime Time) 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 Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool important for the Class 6 board exam?
Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool 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 Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool?
Common mistakes in Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool 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 Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool?
End-of-chapter NCERT exercises for Part 4 — Prime Time Exercises, Summary & Special Numbers Puzzle | Class 6 Maths | MyAiSchool 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 5, and solve at least one previous-year board paper to consolidate your understanding.
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