Step-by-step answers to every "Figure it Out" and in-text question of Chapter 5, Prime Time (NCERT Class 6 Maths, Ganita Prakash, 2026-27): common multiples and common factors (idli-vada and jump games), prime and composite numbers, the Sieve of Eratosthenes, co-prime numbers, prime factorisation, divisibility tests and prime puzzles. All 56 questions are answered, with the key answer highlighted.
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. Multiples of 3 up to 90: 90÷3=30
b. Multiples of 5 up to 90: 90÷5=18
c. Multiples of 15 up to 90: 90÷15=6 (15, 30, 45, 60, 75, 90)
Is this figure (two overlapping circles: multiples of 3 and multiples of 5) somehow related to the 'idli-vada' game? Draw the figure if the game is played till 60.
Solution
Yes. The left circle holds the numbers for which we say 'idli', the right circle those for which we say 'vada', and the overlapping part holds the common multiples, for which we say 'idli-vada'.
Idli-vada up to 60: idli (left), vada (right), idli-vada (middle)
Yes: left part 'idli', right part 'vada', overlap 'idli-vada'. Up to 60, the overlap has 15, 30, 45 and 60.
"Yesterday we played this game with two numbers. We ended up saying just 'idli' or 'idli-vada' 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?
Solution
Nobody said just 'vada' means every multiple of the larger number was also a multiple of the smaller number.
If the other number is 8: idli for 4, vada for 8; every multiple of 8 is a multiple of 4, so 'vada' always comes as 'idli-vada'. ✓
If the other number is 2: idli for 2, vada for 4; every multiple of 4 is a multiple of 2. ✓
3, 5 or 10: no. With 3, 'vada' alone is said on 4 (a multiple of 4 but not of 3); with 5, on 5; with 10, on 10.
8 (and also 2), because every multiple of the larger number is then also a multiple of the smaller one.
In the table of numbers from 31 to 70: 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?
Solution
1. The shaded numbers 33, 36, 39, …, 69 are all multiples of 3.
2. The circled numbers 32, 36, 40, …, 68 are all multiples of 4.
3. 36, 48 and 60 are both shaded and circled. They are common multiples of 3 and 4 (multiples of 12).
Shaded: multiples of 3; circled: multiples of 4; both: 36, 48, 60, the common multiples of 3 and 4.
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, with digit sums 7, 5, 3, 10, 8. Answer: 35.
b. Numbers below 100 with factors 3 and 5 are multiples of 15: 15, 30, 45, 60, 75, 90. Only in 45 does one digit (5) exceed the other (4) by 1.
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 (1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28). Find a perfect number between 1 and 10.
Solution
Factors of 6: 1, 2, 3, 6. Sum = 12 = 2 × 6. So 6 is a perfect number. (Check others: 8 gives 1 + 2 + 4 + 8 = 15, not 16.)
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?
Solution
The first 'idli-vada' is at the first common multiple of the two numbers, so it must be more than 50. Checking pairs below 10: 7 and 8 → 56; 7 and 9 → 63; 8 and 9 → 72. All other pairs have a common multiple of 50 or less (for example 6 and 7 → 42, 5 and 9 → 45).
7 and 8 (first at 56), 7 and 9 (63), or 8 and 9 (72).
In the diagram, Guna has erased all the numbers except the common multiples 24, 48 and 72. Find out what the two numbers could be and fill in the missing numbers in the empty regions.
Solution
The first common multiple is 24, so we need two numbers whose first common multiple is 24, for example 6 and 8 (others: 3 and 8, or 8 and 12).
One possible answer: multiples of 6 and of 8 up to 72 (common multiples 24, 48, 72)
For example 6 and 8: left 6, 12, 18, 30, 36, 42, 54, 60, 66; right 8, 16, 32, 40, 56, 64; middle 24, 48, 72.
Why it works: each composite number has a smaller prime factor, so it gets crossed out as a multiple of that prime; a prime is never a multiple of a smaller number (other than 1), so it is never crossed out. (For numbers up to 100, crossing out multiples of 2, 3, 5 and 7 is enough.)
Are there an equal number of primes occurring in every row in the table (1–10, 11–20, …, 91–100)? Which decades have the least number of primes? Which have the most number of primes?
Solution
Decade
1–10
11–20
21–30
31–40
41–50
51–60
61–70
71–80
81–90
91–100
Primes
4
4
2
2
3
2
2
3
2
1
No, the numbers are not equal. The decade 91–100 has the fewest (only 97); the decades 1–10 and 11–20 have the most (4 each).
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
(5, 7), (11, 13), (29, 31), (41, 43), (59, 61), (71, 73) (besides 3 and 5, and 17 and 19).
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 number ending in 4 is even, so 2 is a factor of it, and it is bigger than 2; it is composite.
b. False. A product of two or more primes has those primes as factors besides 1 and itself (e.g. 2 × 3 = 6), so it is composite.
c. False. Every prime has exactly two factors: 1 and itself.
d. False. 2 is even and prime.
e. True. Every prime after 2 is odd, so the next number is even and bigger than 2, hence composite.
How many three-digit prime numbers can you make using each of 2, 4 and 5 once?
Solution
The numbers are 245, 254, 425, 452, 524, 542. Those ending in 2 or 4 are even (divisible by 2), and those ending in 5 are divisible by 5. So none of them is prime: 0.
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
Prime p
2
5
11
23
29
41
53
2p + 1
5
11
23
47
59
83
107
All of these are primes. (It does not always work: 2 × 7 + 1 = 15 is not prime.)
For example 2 → 5, 5 → 11, 11 → 23, 23 → 47, 29 → 59, 41 → 83.
Anshu observed: 1. Sometimes the first common multiple was the same as the product of the two numbers. 2. At other times the first common multiple was less than the product. Find examples for each. How is it related to the number pair being co-prime?
Solution
Equal to the product: 3 and 5 → 15 = 3 × 5; 4 and 9 → 36 = 4 × 9; 2 and 7 → 14.
Less than the product: 4 and 6 → 12 (product 24); 6 and 8 → 24 (product 48); 10 and 15 → 30 (product 150).
When the two numbers are co-prime, the first common multiple is equal to their product. When they have a common factor (other than 1), the first common multiple is smaller than the product.
Co-prime pairs (3 and 5, 4 and 9): first common multiple = product. Pairs with a common factor (4 and 6, 6 and 8): first common multiple is less than the product.
Co-prime art: In some thread-art diagrams the thread is tied to every peg; in some, it is not. Is it related to the number of pegs and the thread-gap being co-prime? Make such pictures for: a. 15 pegs, thread-gap of 10 b. 10 pegs, thread-gap of 7 c. 14 pegs, thread-gap of 6 d. 8 pegs, thread-gap of 3
Solution
Yes. The thread passes through every peg exactly when the number of pegs and the thread-gap are co-prime. If they share a common factor, the thread returns to the starting peg early and misses some pegs.
Thread art: the thread reaches every peg only when the number of pegs and the thread-gap are co-prime (10 and 7; 8 and 3)
a. 15 and 10 share 5: the thread touches only 15÷5=3 pegs (a triangle).
b. 10 and 7 are co-prime: all 10 pegs (a 10-pointed star).
c. 14 and 6 share 2: only 14÷2=7 pegs (a 7-pointed star).
d. 8 and 3 are co-prime: all 8 pegs (an 8-pointed star).
Yes. All pegs are used only for co-prime pairs: 10 pegs with gap 7 and 8 pegs with gap 3. With 15 and 10 only 3 pegs are used; with 14 and 6 only 7.
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, 45 = 3 × 3 × 5 → common primes 3 and 5: not co-prime.
b. 57 = 3 × 19, 85 = 5 × 17 → no common prime: co-prime.
c. 121 = 11 × 11, 1331 = 11 × 11 × 11 → common prime 11: not co-prime.
d. 343 = 7 × 7 × 7, 216 = 2 × 2 × 2 × 3 × 3 × 3 → no common prime: co-prime.
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 and 231. They have the common prime factors 3 and 7, so they are not co-prime. Neither divides the other: 42 has a factor 2 that 231 does not have, and 231 has a factor 11 that 42 does not have.
Not co-prime (3 and 7 are common factors), and neither divides the other.
Guna says, "Any two prime numbers are co-prime." Is he right?
Solution
Yes, for two different primes. Each prime has only the factors 1 and itself, so two different primes have only 1 as a common factor; for example 5 and 7, or 2 and 13. (A prime is not co-prime with itself: 5 and 5 have the common factor 5.)
Yes: two different primes have no common factor other than 1.
Answer the in-text questions on divisibility: Is 8560 divisible by 10, 5 and 2? What is the largest number less than 399 that is divisible by 5? What are all the multiples of 2 between 399 and 411? Find the numbers divisible by 4 between 330 and 340, 1730 and 1740, and 2030 and 2040. Is 8536 divisible by 4? Find the numbers divisible by 8 between 120 and 140, 1120 and 1140, and 3120 and 3140.
Solution
8560: ends in 0, so it is divisible by 10, 5 and 2.
Largest number below 399 divisible by 5: 395. (125 is not a multiple of 10 because it does not end in 0.)
Multiples of 2 between 399 and 411: 400, 402, 404, 406, 408, 410.
Divisible by 4: 332, 336; 1732, 1736; 2032, 2036. Notice that the last two digits are the same in each group (32 and 36): only the last two digits decide divisibility by 4.
8536: 36 is divisible by 4, so 8536 is divisible by 4.
Divisible by 8: 128, 136; 1128, 1136; 3128, 3136. Here the last three digits decide: 128 and 136 are multiples of 8.
8560 is already a multiple of 8 (560 = 8 × 70). Changing the last two digits, 8504, 8512, 8520, 8528, 8536, 8544, 8552, 8568, 8576, 8584 and 8592 are also multiples of 8.
All the statements in the book are true: for 4, a number is divisible by 4 exactly when the number formed by its last two digits is (because 100 is a multiple of 4); for 8, exactly when its last three digits form a multiple of 8 (because 1000 is a multiple of 8).
8560 is divisible by 10, 5, 2 (and 8); 395; 400 to 410 (even); by 4 only the last two digits matter (332, 336…); by 8 only the last three digits matter (128, 136…).
2024 is a leap year. 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. For a student born in 2014: the leap years till now (2026) are 2016, 2020 and 2024. (Use your own year of birth: list the multiples of 4 from then.)
b. The leap years are 2024, 2028, 2032, …, 2096 (no century year in between). Their number is 42096−2024+1=18+1=19.
a. e.g. born in 2014: 2016, 2020, 2024 b. 19 leap years (2024, 2028, …, 2096).
Explore 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.
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
8 and 5 (or 8 and 10). A number divisible by 8 is also divisible by 4 and 2; a number divisible by both 2 and 5 is divisible by 10.
Within each box (5, 7, 12, 35), (3, 8, 11, 24), (27, 3, 123, 31), (17, 27, 44, 65), try to say how each number is special compared to the rest.
Solution
Box
Possible reasons
5, 7, 12, 35
12: the only even number. 35: the only odd composite number (35 = 5 × 7). 7: the only one that is not a multiple of 2, 3 or 5. 5: the only single-digit multiple of 5.
3, 8, 11, 24
3: the only single-digit odd number. 8: the only cube (2 × 2 × 2) and the only single-digit even number. 11: the only two-digit prime. 24: the only number that is a multiple of all the others except 11 (24 = 3 × 8).
27, 3, 123, 31
3: the only single-digit number. 27: the only cube (3 × 3 × 3). 123: the only 3-digit number. 31: the only number that is not a multiple of 3.
17, 27, 44, 65
17: the only prime. 27: the only multiple of 3 (and a cube). 44: the only even number. 65: the only multiple of 5.
Other reasons are possible; discuss them with your classmates.
For example: 12 the only even, 7 not a multiple of 2, 3 or 5; 11 the only two-digit prime; 31 the only non-multiple of 3; 17 the only prime, 65 the only multiple of 5.
A prime puzzle: Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column. Solve the four puzzles: (i) rows 105, 20, 30; columns 28, 125, 18 (ii) rows 8, 105, 70; columns 30, 70, 28 (iii) rows 63, 27, 190; columns 45, 42, 171 (iv) rows 343, 66, 44; columns 28, 154, 231.
Solution
Method: start with a row or column that has only one possible set of primes (such as 125 = 5 × 5 × 5 or 8 = 2 × 2 × 2), then use the prime factorisations of the other rows and columns.
(i) Column 2 is 125 = 5 × 5 × 5, so it is all 5s. Row 1 then needs 3 × 7 and column 1 (28 = 2 × 2 × 7) has the 7.
7
5
3
105
2
5
2
20
2
5
3
30
28
125
18
(ii) Row 1 is 8 = 2 × 2 × 2. Then column 1 needs 3 × 5, column 3 needs 2 × 7.
2
2
2
8
3
5
7
105
5
7
2
70
30
70
28
(iii) Row 2 is 27 = 3 × 3 × 3. Row 3 is 190 = 2 × 5 × 19, and column 3 (171 = 3 × 3 × 19) must take the 19.
