This MCQ module is based on: 5.4 Prime Factorisation
TOPIC 18 OF 41
5.4 Prime Factorisation
🎓 Class 6
Mathematics
CBSE
Theory
Ch 5 — Prime Time
⏱ ~16 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 5.4 Prime Factorisation
Targeting Class 6 level in Number Theory, with Basic difficulty.
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5.4 Prime Factorisation
Every composite number can be written as a product of primes. Writing a number this way is called prime factorisation?. For example:
\(56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7\)
This is nothing less than the Fundamental Theorem of Arithmetic: every whole number greater than 1 can be written as a product of primes in exactly one way (apart from the order).
Factor Trees
A factor tree breaks a number down step by step into prime factors. Start with the number at the top; at each step, split into two factors. Keep splitting until every leaf is prime.
Factor tree for 72. Red leaves are primes; amber nodes are composite intermediate products.
🔵 Try: Make a factor tree for 60.
60 → 6 × 10 → (2 × 3) × (2 × 5) = \(2^2 \times 3 \times 5\).
60 → 6 × 10 → (2 × 3) × (2 × 5) = \(2^2 \times 3 \times 5\).
Using Prime Factorisation to Check Divisibility
How do we check whether 168 is divisible by 12, without long division? Find prime factorisations:
\(168 = 2 \times 2 \times 2 \times 3 \times 7\)
\(12 = 2 \times 2 \times 3\)
Every prime factor of 12 (with the correct multiplicity) appears inside the factorisation of 168. Therefore 168 is divisible by 12.
Rule
A whole number \(m\) is divisible by \(n\) if and only if every prime factor of \(n\), together with its multiplicity, is present in the prime factorisation of \(m\).
Is 75 divisible by 21?
\(75 = 3 \times 5 \times 5\), \(21 = 3 \times 7\). The prime 7 is missing from 75, so 75 is not divisible by 21.
Check whether 40 and 231 share any common factor (other than 1).
\(40 = 2^3 \cdot 5\); \(231 = 3 \cdot 7 \cdot 11\). No common prime → HCF = 1. They are co-prime.
Find common factors of 242 and 195.
\(242 = 2 \cdot 11^2\); \(195 = 3 \cdot 5 \cdot 13\). No common prime. HCF = 1; co-prime.
5.5 Divisibility Tests
Prime factorisation gives a reliable way to check divisibility, but quick "tests" exist for small divisors.
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 | 346 → last digit 6 → yes |
| 3 | Sum of digits is divisible by 3 | 972 → 9+7+2 = 18 → yes |
| 4 | Number formed by last 2 digits is divisible by 4 | 1,328 → 28/4 = 7 → yes |
| 5 | Last digit is 0 or 5 | 425 → yes |
| 6 | Divisible by both 2 and 3 | 174 → yes |
| 8 | Last 3 digits divisible by 8 | 3,016 → 016/8 = 2 → yes |
| 9 | Sum of digits is divisible by 9 | 2,745 → 2+7+4+5 = 18 → yes |
| 10 | Last digit is 0 | 4,520 → yes |
Why Does the Divisibility Test for 3 Work?
Take the number 246. Write it as \(246 = 2 \cdot 100 + 4 \cdot 10 + 6 = 2(99+1) + 4(9+1) + 6 = (2 \cdot 99 + 4 \cdot 9) + (2 + 4 + 6)\). The first bracket is already divisible by 3 (both 99 and 9 are). So 246 is divisible by 3 if and only if the sum of digits (2 + 4 + 6 = 12) is divisible by 3. The same argument works for 9 since 99 and 9 are divisible by 9.
Why Does the Divisibility Test for 4 Work?
Any number can be written as (hundreds × 100) + (last two digits). Since 100 is a multiple of 4, the "hundreds × 100" part is always divisible by 4. So divisibility by 4 depends only on the last two digits. Similarly, divisibility by 8 depends on the last three digits because 1000 = 8 × 125.
🔵 Change the last two digits of 8560 so that the resulting number is a multiple of 8. The thousand-part, 8000, is already a multiple of 8. So we need the last three digits \(560 + \Delta\) to be divisible by 8. Replace 560 by 552 → 8552 works.
Leap Year Check — Divisibility by 4 and 400
A leap year is divisible by 4, unless it is a century year — in which case it must also be divisible by 400. So 2024 is a leap year (div. by 4); 1900 is NOT (div. by 4 but a century year not div. by 400); 2000 IS (div. by 400).
How many years between 1999 and 2099 are leap years?
Multiples of 4 from 2000 to 2096: 2000, 2004, …, 2096 → 25 numbers. 2100 is a century not divisible by 400 (excluded). 2000 is divisible by 400 (kept). So 25 leap years between 1999 and 2099.
Figure it Out (Section 5.5)
Q1. Change the last two digits of 8560 so that it becomes a multiple of 8.
8552 is a multiple of 8 (552 = 8 × 69). Many possibilities.
Q2. State whether these are True or False:
(1) If the last three digits matter when deciding divisibility by 8, then dividing the number formed by the last three digits by 8 gives the original number divisible by 8.
(2) If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
(3) If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
(1) If the last three digits matter when deciding divisibility by 8, then dividing the number formed by the last three digits by 8 gives the original number divisible by 8.
(2) If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
(3) If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
(1) True. (2) True. (3) True. Examples: 8576, 7648, 5024.
Q3. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Largest: 9999 — last 2 digits 99 not divisible by 4. Try 8888 → 88/4 = 22 → works. Largest = 8888. Smallest: 1001 → 01 no. 1111, 1221… check 1001+? Palindromes 4-digit form ABBA. Need BA divisible by 4. Smallest ABBA with BA divisible by 4: A=1, BA must be divisible by 4, so B=0 gives "01"=1 no; B=2 "21" no; B=4 "41" no; B=6 "61" no; B=8 "81" no. Try A=2, BA=B2: B=1→12 yes → 2112. So smallest = 1001… actually re-check: the palindrome 4-digit number ABBA has last two digits "BA". For A=1, BA ends in 1 (odd) — never divisible by 4. For A=2, BA=B2 → B=1 gives 12 (÷4 yes) → 2112. Smallest = 2112.
Q4. Explain and find out if each statement is always true, sometimes true or never true:
(a) Sum of two even numbers gives a multiple of 4.
(b) Sum of two odd numbers gives a multiple of 4.
(a) Sum of two even numbers gives a multiple of 4.
(b) Sum of two odd numbers gives a multiple of 4.
(a) Sometimes true: 2+2=4 ✓, 2+4=6 ✗. (b) Sometimes true: 1+3=4 ✓, 1+5=6 ✗.
Q5. Find remainders when each of 78, 99, 173, 572, 980, 1111, 2345 is divided by (i) 10, (ii) 5, (iii) 2.
| N | ÷10 | ÷5 | ÷2 |
|---|---|---|---|
| 78 | 8 | 3 | 0 |
| 99 | 9 | 4 | 1 |
| 173 | 3 | 3 | 1 |
| 572 | 2 | 2 | 0 |
| 980 | 0 | 0 | 0 |
| 1111 | 1 | 1 | 1 |
| 2345 | 5 | 0 | 1 |
Q6. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked only two and declared that it was also divisible by all. What could those two numbers be?
If divisible by 8 and 10, then also by 2, 4, 5.
Q7. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160?
Divisible by 2, 4, 5, 8, 10 means divisible by LCM(2,4,5,8,10) = 40. 572/40 no. 2352/40 no. 5600/40 yes. 6000/40 yes. 77622160/40 yes. Answer: 5600, 6000, 77622160.
Q8. Write two numbers whose product is 10000. The two numbers should not have 0 as their unit digit.
\(10000 = 2^4 \cdot 5^4 = 16 \cdot 625\). Neither 16 nor 625 ends in 0. 16 × 625 = 10000.
Activity: Divisibility Detective
L3 Apply
Materials: Calendar of current year, notebook
Predict: Without looking, how many days in this year are in months whose number is divisible by 3? (i.e. Mar, Jun, Sep, Dec.)
- List the month-numbers 1 to 12. Mark the ones divisible by 3.
- Add up the days in each of those months.
- Check your prediction. Repeat for months divisible by 4 (Apr, Aug, Dec).
- Extension: list all dates (dd/mm) in the year where BOTH dd and mm are divisible by 3.
Mar+Jun+Sep+Dec = 31+30+30+31 = 122 days.
Competency-Based Questions
Scenario: A cash-counter in a canteen has the number 7,23,456 as the day's collection slip. The manager needs to know quickly, without a calculator, whether this amount can be split equally among 2, 3, 4, 5, 6, 8, or 9 employees.
Q1. Which of the divisors 2, 3, 4, 5 divide 7,23,456 evenly?
L3 ApplyLast digit 6 → div by 2 yes. Sum of digits 7+2+3+4+5+6 = 27 → div by 3 yes. Last two digits 56 → 56/4 = 14 yes. Last digit not 0 or 5 → not div by 5. So divisible by 2, 3, 4 (not 5).
Q2. Is it divisible by 6, by 8, and by 9? Justify using the rules.
L4 Analyse6: div by both 2 and 3 → yes. 8: last three digits 456, 456/8 = 57 → yes. 9: digit sum 27, 27/9 = 3 → yes.
Q3. Gauri claims: "Any 6-digit number with digits 1, 2, 3, 4, 5, 6 (in some order) is divisible by 3." Evaluate and justify.
L5 EvaluateSum of digits 1+2+3+4+5+6 = 21, divisible by 3 irrespective of order. Claim is TRUE. In fact, divisibility by 3 depends only on the digit sum, not order.
Q4. Design the smallest 5-digit number that is divisible by 4, 9 and 25 simultaneously. Show your reasoning.
L6 CreateLCM(4, 9, 25) = 900. Smallest 5-digit multiple of 900: ⌈10000/900⌉ = 12, so 12 × 900 = 10800. Check: 10800/4 = 2700 ✓, 10800/9 = 1200 ✓, 10800/25 = 432 ✓.
Assertion–Reason Questions
A: A number is divisible by 9 if the sum of its digits is divisible by 9.
R: 99, 9999, 999 999… are all divisible by 9.
R: 99, 9999, 999 999… are all divisible by 9.
Answer: (a) — the rule works because powers of 10 minus 1 are always multiples of 9.
A: 1000 is divisible by 8.
R: Divisibility by 8 depends on the last three digits of a number.
R: Divisibility by 8 depends on the last three digits of a number.
Answer: (b) — both true. 1000 = 8 × 125. R is also true (since 1000 is a multiple of 8). R is the reason the divisibility-by-8 RULE works, but doesn't specifically explain why 1000 is divisible by 8.
A: Every number divisible by 6 is divisible by both 2 and 3.
R: 6 = 2 × 3 and 2 and 3 are co-prime.
R: 6 = 2 × 3 and 2 and 3 are co-prime.
Answer: (a) — any multiple of 6 contains both primes 2 and 3 in its factorisation.
Frequently Asked Questions — Prime Time
What is Part 3 — Prime Factorisation & Divisibility Tests | Class 6 Maths | MyAiSchool in NCERT Class 6 Mathematics?
Part 3 — Prime Factorisation & Divisibility Tests | 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 3 — Prime Factorisation & Divisibility Tests | Class 6 Maths | MyAiSchool step by step?
To solve problems on Part 3 — Prime Factorisation & Divisibility Tests | 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 3 — Prime Factorisation & Divisibility Tests | Class 6 Maths | MyAiSchool important for the Class 6 board exam?
Part 3 — Prime Factorisation & Divisibility Tests | 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 3 — Prime Factorisation & Divisibility Tests | Class 6 Maths | MyAiSchool?
Common mistakes in Part 3 — Prime Factorisation & Divisibility Tests | 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 3 — Prime Factorisation & Divisibility Tests | Class 6 Maths | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — Prime Factorisation & Divisibility Tests | 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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