Class 6 Maths Ganita Prakash Chapter 5 Solutions (NCERT 2026–27) – Prime Time

These Class 6 Maths Ganita Prakash Chapter 5 solutions cover Prime Time from the new NCF textbook (Reprint 2026–27). Every Figure it Out, Math Talk and Try This task is solved step by step — with common multiples, common factors, prime and composite numbers, prime factorisation, co-prime numbers and divisibility tests — so you can master the chapter and revise it quickly.

Class: 6 Subject: Mathematics Book: Ganita Prakash Chapter: 5 Exercises: Figure it Out (5.1, 5.2, 5.4, 5.5) + Math Talk & Try This Session: 2026–27
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Chapter 5 Overview

Chapter 5 of Ganita Prakash, Prime Time, uses playful games — the Idli-Vada game and Grumpy and Jumpy’s treasure hunt — to introduce common multiples and common factors. It then builds the ideas of prime and composite numbers (with the Sieve of Eratosthenes), co-prime numbers, prime factorisation and how it is used to test co-primeness and divisibility, and the quick divisibility tests for 10, 5, 2, 4 and 8. The Class 6 Maths Ganita Prakash Chapter 5 solutions below work through every Figure it Out, Math Talk and Try This question step by step.

Key Concepts & Definitions

Multiple: the result of multiplying a number by 1, 2, 3, … (e.g. multiples of 3 are 3, 6, 9, 12, …). A common multiple of two numbers is a multiple of both (e.g. 15 is a common multiple of 3 and 5).

Factor (divisor): a number that divides another exactly with no remainder (e.g. the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24). A common factor divides both numbers (e.g. 1 and 2 are the common factors of 14 and 36).

Prime number: a number that has exactly two factors — 1 and itself (e.g. 2, 3, 5, 7, 11, …). 2 is the only even prime.

Composite number: a number with more than two factors (e.g. 4, 6, 8, 9, …). The number 1 is neither prime nor composite.

Co-prime numbers: two numbers that have no common factor other than 1 (e.g. 4 and 9). They need not be prime themselves.

Prime factorisation: writing a number as a product of prime numbers only (e.g. 56 = 2 × 2 × 2 × 7). Every number greater than 1 has exactly one prime factorisation, apart from the order of factors.

Perfect number: a number whose factors (including itself) add up to twice the number (e.g. 6 and 28).

Important Rules & Tests (Chapter 5)

Co-prime test: two numbers are co-prime when their prime factorisations share no common prime factor.

Divisibility test (using prime factorisation): a number is divisible by another when the prime factorisation of the second is fully included in that of the first.

Divisible by 10: the number ends in 0.

Divisible by 5: the number ends in 0 or 5.

Divisible by 2: the number ends in 0, 2, 4, 6 or 8 (it is even).

Divisible by 4: the number formed by the last two digits is divisible by 4.

Divisible by 8: the number formed by the last three digits is divisible by 8.

Figure it Out — 5.1 Common Multiples and Common Factors (Page 108 & 110–111)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified against the answers given in the book.

Figure it Out (Page 108)

1. At what number is ‘idli-vada’ said for the 10th time?

SOLUTION ‘Idli-vada’ is said at common multiples of 3 and 5, i.e. multiples of 15: 15, 30, 45, … The 10th such number is 15 × 10 = 150.

2. If the game is played for the numbers 1 to 90, find out: a. How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)? b. How many times would the children say ‘vada’ (including the times they say ‘idli-vada’)? c. How many times would the children say ‘idli-vada’?

SOLUTION a. ‘Idli’ is said for every multiple of 3 up to 90 → 90 ÷ 3 = 30 times. b. ‘Vada’ is said for every multiple of 5 up to 90 → 90 ÷ 5 = 18 times. c. ‘Idli-vada’ is said for every multiple of 15 up to 90 → 90 ÷ 15 = 6 times.

3. What if the game was played till 900? How would your answers change?

SOLUTION ‘Idli’ (multiples of 3): 900 ÷ 3 = 300 times. ‘Vada’ (multiples of 5): 900 ÷ 5 = 180 times. ‘Idli-vada’ (multiples of 15): 900 ÷ 15 = 60 times.

4. Is this figure somehow related to the ‘idli-vada’ game? Hint: Imagine playing the game till 30. Draw the figure if the game is played till 60.

SOLUTION Yes. The figure is two overlapping circles — one for multiples of 3, one for multiples of 5. The numbers in the overlap (15, 30, …) are the common multiples, which are exactly the numbers where the players say ‘idli-vada’. For the game till 60 the overlap would contain 15, 30, 45 and 60.

Figure it Out (Page 110–111)

1. Find all multiples of 40 that lie between 310 and 410.

SOLUTION Multiples of 40 are 40, 80, …, 280, 320, 360, 400, 440, … Those between 310 and 410 are 320, 360 and 400.

2. Who am I? a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8. b. I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.

SOLUTION a. Multiples of 7 below 40: 7, 14, 21, 28, 35. The one whose digits add to 8 is 35 (3 + 5 = 8). ∴ 35. b. The number is a multiple of both 3 and 5, so a multiple of 15 below 100: 15, 30, 45, 60, 75, 90. The one whose digits differ by 1 is 45 (5 is 1 more than 4). ∴ 45.

3. A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.

SOLUTION Try 6: its factors are 1, 2, 3 and 6. Their sum = 1 + 2 + 3 + 6 = 12 = 2 × 6. ∴ the perfect number between 1 and 10 is 6.

4. Find the common factors of: a. 20 and 28    b. 35 and 50 c. 4, 8 and 12    d. 5, 15 and 25

SOLUTION a. Factors of 20 = 1, 2, 4, 5, 10, 20; factors of 28 = 1, 2, 4, 7, 14, 28. Common factors = 1, 2, 4. b. Factors of 35 = 1, 5, 7, 35; factors of 50 = 1, 2, 5, 10, 25, 50. Common factors = 1, 5. c. Common factors of 4, 8 and 12 = 1, 2, 4. d. Common factors of 5, 15 and 25 = 1, 5.

5. Find any three numbers that are multiples of 25 but not multiples of 50.

SOLUTION Multiples of 25 are 25, 50, 75, 100, 125, 150, … The odd multiples of 25 are not multiples of 50. Three such numbers are 25, 75 and 125. (Other valid answers: 175, 225, …)

6. Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both smaller than 10. The first time anybody says ‘idli-vada’ is after the number 50. What could the two numbers be which are assigned ‘idli’ and ‘vada’?

SOLUTION The first ‘idli-vada’ is the first common multiple of the two numbers, and it must be greater than 50. For co-prime numbers below 10 the first common multiple is their product. 7 × 8 = 56 (> 50) and 8 × 9 = 72 (> 50), both first common multiples beyond 50. ∴ the two numbers could be 7 and 8, or 8 and 9. (Try for other possibilities.)

7. In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both the numbers?

SOLUTION A jump size lands on both numbers only if it is a common factor of 28 and 70. Factors of 28 = 1, 2, 4, 7, 14, 28; factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70. ∴ the jump sizes are 1, 2, 7 and 14.

8. In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions. (Common multiples shown: 24, 48, 72.)

SOLUTION The common multiples 24, 48 and 72 are all multiples of 8 and of 3 (and of 12, 6, etc.). A neat choice is multiples of 3 and multiples of 8, whose common multiples are 24, 48, 72. Then the ‘multiples of 3 only’ region could be filled with 3, 6, 9, … and the ‘multiples of 8 only’ region with 8, 16, 32, … Other answers are possible, e.g. multiples of 6 and 8, or 12 and 8 — any pair whose common multiples are 24, 48, 72.

9. Find the smallest number that is a multiple of all the numbers from 1 to 10, except for 7.

SOLUTION We need the LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10. Highest prime powers: 23 (from 8), 32 (from 9), 5 (from 5/10). LCM = 23 × 32 × 5 = 8 × 9 × 5 = 360.

10. Find the smallest number that is a multiple of all the numbers from 1 to 10.

SOLUTION Now include 7 as well: LCM = 23 × 32 × 5 × 7 = 360 × 7 = 2520.

Figure it Out — 5.2 Prime Numbers (Page 114)

1. We see that 2 is a prime and also an even number. Is there any other even prime?

SOLUTION No. Every even number other than 2 is divisible by 2, so it has at least three factors (1, 2 and itself) and is composite. Hence 2 is the only even prime.

2. Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference?

SOLUTION The smallest difference is 1, between the only consecutive primes 2 and 3 (3 − 2 = 1). The largest gap up to 100 is 8, between 89 and 97 (97 − 89 = 8).

3. Are there an equal number of primes occurring in every row in the table on the previous page? Which decades have the least number of primes? Which have the most number of primes?

SOLUTION No, the rows do not have equal numbers of primes. The decade with the least is 91–100 (only 97 → 1 prime). The decades with the most are 1–10 and 11–20 (4 primes each: 2, 3, 5, 7 and 11, 13, 17, 19).

4. Which of the following numbers are prime: 23, 51, 37, 26?

SOLUTION 23 → only factors 1 and 23 → prime. 51 = 3 × 17 → composite. 37 → only factors 1 and 37 → prime. 26 = 2 × 13 → composite. ∴ the primes are 23 and 37.

5. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.

SOLUTION (2, 3) → 2 + 3 = 5; (3, 7) → 3 + 7 = 10; (2, 13) → 2 + 13 = 15 — each sum is a multiple of 5. ∴ three pairs are (2, 3), (3, 7) and (2, 13). (Other pairs, e.g. (7, 13) = 20, also work.)

6. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.

SOLUTION We need primes below 100 that stay prime when their digits are reversed: 17 and 71, 37 and 73, 79 and 97 (besides the given 13 and 31).

7. Find seven consecutive composite numbers between 1 and 100.

SOLUTION Between the primes 89 and 97 there are seven numbers in a row, none of them prime: 90, 91, 92, 93, 94, 95, 96 (note 91 = 7 × 13). All are composite.

8. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.

SOLUTION All twin-prime pairs up to 100 are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).

9. Identify whether each statement is true or false. Explain. a. There is no prime number whose units digit is 4. b. A product of primes can also be prime. c. Prime numbers do not have any factors. d. All even numbers are composite numbers. e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite.

SOLUTION a. True. A units digit of 4 means the number is even, so it is divisible by 2; the only even prime is 2, which does not end in 4. So no prime ends in 4. b. False. A product of two or more primes has those primes as factors, so it is composite (e.g. 2 × 3 = 6). c. False. Every prime has exactly two factors — 1 and the number itself. d. False. All even numbers except 2 are composite; 2 is an even prime. e. True. Except for the pair (2, 3), any prime > 2 is odd, so the next number is even (and greater than 2), hence composite.

10. Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330?

SOLUTION 45 = 3 × 3 × 5 (only two distinct primes); 60 = 2 × 2 × 3 × 5 (a prime repeats); 91 = 7 × 13 (only two primes); 330 = 2 × 3 × 5 × 11 (four primes). 105 = 3 × 5 × 7 — exactly three distinct primes. ∴ the answer is 105.

11. How many three-digit prime numbers can you make using each of 2, 4 and 5 once?

SOLUTION The six arrangements are 245, 254, 425, 452, 524, 542. Any number ending in 2 or 4 is even, and any ending in 5 is divisible by 5 — so all six are composite. ∴ the number of three-digit primes is None (0).

12. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples.

SOLUTION Take a prime p and check if 2p + 1 is prime: p = 5 → 2 × 5 + 1 = 11 (prime); p = 11 → 23 (prime); p = 23 → 47 (prime); p = 29 → 59 (prime); p = 41 → 83 (prime). ∴ five examples: 5→11, 11→23, 23→47, 29→59, 41→83. (Try other possibilities.)

Figure it Out — 5.4 Prime Factorisation (Page 120 & 122)

Figure it Out (Page 120)

1. Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.

SOLUTION 64 = 2 × 2 × 2 × 2 × 2 × 2 (= 26) 104 = 2 × 2 × 2 × 13 105 = 3 × 5 × 7 243 = 3 × 3 × 3 × 3 × 3 (= 35) 320 = 2 × 2 × 2 × 2 × 2 × 2 × 5 (= 26 × 5) 141 = 3 × 47 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 (= 26 × 33) 729 = 3 × 3 × 3 × 3 × 3 × 3 (= 36) 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (= 210) 1331 = 11 × 11 × 11 (= 113) 1000 = 2 × 2 × 2 × 5 × 5 × 5 (= 23 × 53)

2. The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?

SOLUTION Number = 2 × 3 × 3 × 11 = 2 × 9 × 11 = 198.

3. Find three prime numbers, all less than 30, whose product is 1955.

SOLUTION 1955 ends in 5, so divide by 5: 1955 ÷ 5 = 391. Now 391 = 17 × 23. ∴ 1955 = 5 × 17 × 23 (all primes less than 30).

4. Find the prime factorisation of these numbers without multiplying first. a. 56 × 25    b. 108 × 75    c. 1000 × 81

SOLUTION Just join the prime factorisations of each factor. a. 56 = 2 × 2 × 2 × 7 and 25 = 5 × 5, so 56 × 25 = 2 × 2 × 2 × 5 × 5 × 7. b. 108 = 2 × 2 × 3 × 3 × 3 and 75 = 3 × 5 × 5, so 108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5. c. 1000 = 2 × 2 × 2 × 5 × 5 × 5 and 81 = 3 × 3 × 3 × 3, so 1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5.

5. What is the smallest number whose prime factorisation has: a. three different prime numbers? b. four different prime numbers?

SOLUTION Use the smallest primes, each once. a. 2 × 3 × 5 = 30. b. 2 × 3 × 5 × 7 = 210.

Figure it Out (Page 122)

1. Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer. a. 30 and 45    b. 57 and 85 c. 121 and 1331    d. 343 and 216

SOLUTION a. 30 = 2 × 3 × 5 and 45 = 3 × 3 × 5. Common primes 3 and 5 → not co-prime. b. 57 = 3 × 19 and 85 = 5 × 17. No common prime → co-prime. c. 121 = 11 × 11 and 1331 = 11 × 11 × 11. Common prime 11 → not co-prime. d. 343 = 7 × 7 × 7 and 216 = 2 × 2 × 2 × 3 × 3 × 3. No common prime → co-prime.

2. Is the first number divisible by the second? Use prime factorisation. a. 225 and 27    b. 96 and 24 c. 343 and 17    d. 999 and 99

SOLUTION a. 225 = 3 × 3 × 5 × 5 and 27 = 3 × 3 × 3. The three 3s of 27 are not all in 225 (only two 3s) → No. b. 96 = 2 × 2 × 2 × 2 × 2 × 3 and 24 = 2 × 2 × 2 × 3. All factors of 24 appear in 96 → Yes. c. 343 = 7 × 7 × 7 and 17 = 17. 17 is not a factor of 343 → No. d. 999 = 3 × 3 × 3 × 37 and 99 = 3 × 3 × 11. 11 is not a factor of 999 → No.

3. The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?

SOLUTION The numbers are 42 (= 2 × 3 × 7) and 231 (= 3 × 7 × 11). They share the primes 3 and 7, so they are not co-prime. 231 has the prime 11 (absent in 42), and 42 has the prime 2 (absent in 231), so neither divides the other.

4. Guna says, “Any two prime numbers are co-prime”. Is he right?

SOLUTION Yes. A prime has only itself and 1 as factors, so two different primes can share no factor except 1 (e.g. 2 and 3, or 3 and 11). Hence any two distinct primes are co-prime.

Figure it Out — 5.5 Divisibility Tests (Page 125)

1. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. a. From the year you were born till now, which years were leap years? b. From the year 2024 till 2099, how many leap years are there?

SOLUTION a. This depends on your birth year — list the multiples of 4 in that range. For example, a student born in 2014 saw the leap years 2016, 2020 and 2024. b. Leap years from 2024 to 2099 are the multiples of 4: 2024, 2028, …, 2096. Count = (2096 − 2024) ÷ 4 + 1 = 72 ÷ 4 + 1 = 19. (None of these is a century year, so none is excepted.)

2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.

SOLUTION A 4-digit palindrome has the form “abba”; for divisibility by 4 the last two digits “ba” must be divisible by 4. Largest: 8888 (last two digits 88 = 4 × 22). (The book’s answer key lists 9999; however 9999 is odd and not divisible by 4, so the correct largest palindrome divisible by 4 is 8888.) Smallest: 2112 (last two digits 12 = 4 × 3). (The book lists 1001, but 1001 is odd and not divisible by 4; the smallest 4-digit palindrome divisible by 4 is 2112.)

3. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning. a. Sum of two even numbers gives a multiple of 4. b. Sum of two odd numbers gives a multiple of 4.

SOLUTION a. Sometimes true. 2 + 6 = 8 (a multiple of 4), but 2 + 4 = 6 (not a multiple of 4). b. Sometimes true. 1 + 3 = 4 (a multiple of 4), but 1 + 5 = 6 (not a multiple of 4).

4. Find the remainders obtained when each of the following numbers are divided by (a) 10, (b) 5, (c) 2. 78, 99, 173, 572, 980, 1111, 2345

SOLUTION Remainder by 10 = last digit; by 5 = last digit mod 5; by 2 = 0 if even, 1 if odd.
Number÷ 10÷ 5÷ 2
78830
99941
173331
572220
980000
1111111
2345501

5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?

SOLUTION If a number is divisible by 8 it is also divisible by 4 and 2; if it is divisible by 5 and by 2 it is divisible by 10. So checking the two ‘biggest’ conditions covers all the rest. ∴ the two numbers are 5 and 8 (14560 ends in 0 → divisible by 5, and its last three digits 560 = 8 × 70 → divisible by 8).

6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160.

SOLUTION A number divisible by all of 2, 4, 5, 8 and 10 must end in 0 (for 5 and 10) and have its last three digits divisible by 8. 572 → ends in 2, not divisible by 5/10 → No. 2352 → ends in 2 → No. 5600 → ends in 0, last three digits 600 = 8 × 75 → Yes. 6000 → ends in 0, 000 divisible by 8 → Yes. 77622160 → ends in 0, last three digits 160 = 8 × 20 → Yes. ∴ 5600, 6000 and 77622160.

7. Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.

SOLUTION 10000 = 24 × 54. To avoid a units digit of 0 (which needs both a 2 and a 5), put all the 2s in one number and all the 5s in the other: 24 = 16 and 54 = 625. ∴ 16 × 625 = 10000, and neither 16 nor 625 ends in 0.

Math Talk & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

In-text (Page 109) — The mystery number pair Yesterday we played with two numbers and nobody said just ‘vada’. One of the numbers was 4. Which of the following could be the other number: 2, 3, 5, 8, 10? Answer. 8. Nobody saying just ‘vada’ means every multiple of the larger number is also a multiple of 4 (so it always becomes ‘idli’ or ‘idli-vada’). This happens when the larger number is itself a multiple of 4. Among the options, only 8 is a multiple of 4.
In-text (Page 110) — Jump sizes for 15 and 30 What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. Answer. A working jump size must be a common factor of 15 and 30. Factors of 15 = 1, 3, 5, 15; factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30. Common factors are 1, 3, 5 and 15.
Math Talk (Page 110) — The shaded/circled table 1. Is there anything common among the shaded numbers? 2. Is there anything common among the circled numbers? 3. Which numbers are both shaded and circled? What are these numbers called? Answer. 1. All shaded numbers are multiples of 3. 2. All circled numbers are multiples of 4. 3. The numbers that are both shaded and circled are 36, 48 and 60; these are the common multiples of 3 and 4.
In-text (Page 113) — Primes and composites from 21 to 30 How many prime numbers are there from 21 to 30? How many composite numbers are there from 21 to 30? Answer. Primes from 21 to 30: 2 of them (23 and 29). Composite numbers from 21 to 30: 8 of them (21, 22, 24, 25, 26, 27, 28, 30).
In-text (Page 115) — Safe treasure pairs Check if these pairs are safe (Jumpy cannot reach both with any jump size other than 1): a. 15 and 39, b. 4 and 15, c. 18 and 29, d. 20 and 55. Answer. A pair is safe only when the two numbers are co-prime. 15 & 39 share 3 (not safe); 4 & 15 share only 1 (safe); 18 & 29 share only 1 (safe); 20 & 55 share 5 (not safe). So the safe pairs are 4 & 15 and 18 & 29.
In-text (Page 116) — Which pairs are co-prime? Which of the following pairs of numbers are co-prime? a. 18 and 35, b. 15 and 37, c. 30 and 415, d. 17 and 69, e. 81 and 18. Answer. a. 18 = 2×3×3, 35 = 5×7 → co-prime. b. 15 = 3×5, 37 prime → co-prime. c. 30 = 2×3×5, 415 = 5×83 → share 5, not co-prime. d. 17 prime, 69 = 3×23 → co-prime. e. 81 = 34, 18 = 2×3×3 → share 3, not co-prime. So the co-prime pairs are a, b and d.
Math Talk (Page 116) — First common multiple vs product Sometimes the first common multiple equals the product of the two numbers; at other times it is less. Find examples for each. How is it related to the pair being co-prime? Answer. Equal to the product (co-prime pairs): 3 & 5 (first common multiple 15), 3 & 7 (21), 4 & 9 (36). Less than the product (not co-prime): 3 & 6 (first common multiple 6, not 18), 3 & 12 (12), 6 & 15 (30). When the pair is co-prime, the first common multiple equals the product; otherwise it is less.
Math Talk (Page 123–124) — Divisibility statements Do you agree that a number is divisible by 10 if it ends in 0; by 5 if it ends in 0 or 5; by 2 if it ends in 0, 2, 4, 6 or 8? Answer. Yes to all three. So 8560 is divisible by 10, 5 and 2 (it ends in 0). The multiples of 2 between 399 and 411 are 400, 402, 404, 406, 408 and 410.
Math Talk (Page 124) — Divisibility by 4 Is 8536 divisible by 4? Do you agree that only the last two digits matter, and that the number is divisible by 4 exactly when the last two digits are? Answer. Yes — 8536 is divisible by 4 because its last two digits, 36 = 4 × 9, are divisible by 4. All three statements about ‘last two digits’ are correct (e.g. 124, 364, 4028 are all divisible by 4).
Math Talk (Page 125) — Divisibility by 8 Find numbers between 120–140, 1120–1140 and 3120–3140 that are divisible by 8. Change the last two digits of 8560 to make a multiple of 8. Do you agree the last three digits decide divisibility by 8? Answer. 120–140: 128 and 136; 1120–1140: 1128 and 1136; 3120–3140: 3128 and 3136. We observe the same last-two-digit pattern. Changing 8560 to 8552 gives a multiple of 8 (552 = 8 × 69). Yes — the last three digits decide divisibility by 8 (e.g. 8576, 7648, 5024).
Try This (Page 116) — Co-prime thread art Make thread-art pictures for: a. 15 pegs, gap 10; b. 10 pegs, gap 7; c. 14 pegs, gap 6; d. 8 pegs, gap 3. When is the thread tied to every peg? Answer. The thread touches every peg exactly when the number of pegs and the thread-gap are co-prime. b. 10 & 7 are co-prime → every peg is used; d. 8 & 3 are co-prime → every peg is used. a. 15 & 10 share 5 → not every peg; c. 14 & 6 share 2 → not every peg.

Common Mistakes to Avoid

Watch out for these

  • Calling 1 a prime — the number 1 is neither prime nor composite (it has only one factor).
  • Thinking co-prime numbers must themselves be prime — e.g. 4 and 9 are co-prime but neither is prime.
  • Mixing up factors and multiples: a jump size that reaches a treasure is a factor; the idli-vada numbers are common multiples.
  • Concluding two numbers are co-prime from one factorisation like 56 = 14 × 4 — always use the full prime factorisation.
  • Testing divisibility by 4 with the last one digit — use the last two digits for 4 and the last three digits for 8.
  • Forgetting that the only even prime is 2 — every other even number is composite.

Practice MCQs & Assertion–Reason

1. The smallest prime number is:

(a) 0    (b) 1    (c) 2    (d) 3

2. Which number is neither prime nor composite?

(a) 1    (b) 2    (c) 3    (d) 4

3. The common factors of 20 and 28 are:

(a) 1, 2    (b) 1, 2, 4    (c) 2, 4, 5    (d) 1, 4, 7

4. Which of these pairs is co-prime?

(a) 12 and 18    (b) 15 and 39    (c) 4 and 15    (d) 20 and 55

5. The prime factorisation of 105 is:

(a) 3 × 5 × 7    (b) 5 × 21    (c) 3 × 35    (d) 5 × 5 × 3

6. The smallest number that is a multiple of all numbers from 1 to 10 is:

(a) 360    (b) 840    (c) 1260    (d) 2520

7. A number is divisible by 8 if the number formed by its last:

(a) one digit is divisible by 8    (b) two digits is divisible by 8    (c) three digits is divisible by 8    (d) four digits is divisible by 8

8. Which of the following is the product of exactly three distinct primes?

(a) 45    (b) 60    (c) 91    (d) 105

9. A perfect number between 1 and 10 is:

(a) 4    (b) 6    (c) 8    (d) 9

10. Which pair is a set of twin primes?

(a) 7 and 11    (b) 13 and 17    (c) 17 and 19    (d) 23 and 29

Answer key: 1-(c), 2-(a), 3-(b), 4-(c), 5-(a), 6-(d), 7-(c), 8-(d), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 2 is the only even prime number.

Reason: Every even number greater than 2 is divisible by 2, so it has more than two factors.

A-R 2. Assertion: The numbers 4 and 9 are co-prime.

Reason: Two numbers are co-prime only when both of them are prime numbers.

A-R 3. Assertion: 1 is a prime number.

Reason: A prime number has exactly two factors, 1 and itself.

A-R 4. Assertion: 105 = 3 × 5 × 7 is the prime factorisation of 105.

Reason: Every number greater than 1 has exactly one prime factorisation, apart from the order of factors.

A-R 5. Assertion: 5600 is divisible by 8.

Reason: The number formed by its last three digits, 600, is divisible by 8.

Answer key: 1-(A), 2-(C), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • A common multiple of two numbers is a multiple of both; a common factor divides both.
  • A prime number has exactly two factors; a composite number has more than two; 1 is neither.
  • 2 is the only even prime; the Sieve of Eratosthenes lists primes up to 100.
  • Two numbers are co-prime if their only common factor is 1 (they need not be prime).
  • Prime factorisation writes a number as a product of primes, and it is unique apart from order.
  • Use prime factorisation to test co-primeness (no shared prime) and divisibility (second’s primes contained in the first’s).
  • Divisibility: ends in 0 (by 10); ends in 0/5 (by 5); even (by 2); last two digits (by 4); last three digits (by 8).

How to score full marks in this chapter

Always write the full prime factorisation before deciding if two numbers are co-prime or whether one divides the other — never judge from a single split like 56 = 14 × 4. For divisibility, quote the exact rule you are using (“last two digits for 4”, “last three for 8”). For ‘smallest number’ questions take the highest power of each prime (LCM). Keep factor lists and multiple lists clearly separated so you do not confuse jump sizes (factors) with idli-vada numbers (common multiples).

Frequently Asked Questions

What is Class 6 Maths Ganita Prakash Chapter 5 about?

Chapter 5, Prime Time, covers common multiples and common factors (through the Idli-Vada and treasure-hunt games), prime and composite numbers, the Sieve of Eratosthenes, co-prime numbers, prime factorisation and its uses, and the divisibility tests for 10, 5, 2, 4 and 8.

How many Figure it Out exercises are there in Chapter 5?

There are four main “Figure it Out” sets — in Section 5.1 (Common Multiples and Common Factors), 5.2 (Prime Numbers), 5.4 (Prime Factorisation) and 5.5 (Divisibility Tests) — plus several Math Talk and Try This tasks, all solved on this page.

Why is 1 neither a prime nor a composite number?

A prime has exactly two factors and a composite has more than two. The number 1 has only one factor (itself), so it fits neither definition — that is why 1 is classed as neither prime nor composite.

Are these Class 6 Maths Ganita Prakash Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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