Chapter 5 : Prime Time (NCERT Solution)

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Pick a section, then choose In-text questions (asked within the lesson) or Figure it Out (the exercises).

5.1 Common Multiples & Common Factors

In-text Questions Figure it Out

5.2 Prime Numbers

In-text Questions Figure it Out

5.3 Co-prime Numbers

In-text Questions

5.4 Prime Factorisation

Figure it Out

5.5 Divisibility Tests

In-text Questions Figure it Out

5.1

Common Multiples & Common Factors

In-text Questions asked within the lesson

The children played the game with two numbers and ended up saying only ‘idli’ or ‘idli-vada’, and nobody said just ‘vada’. One of the numbers was 4. Which of these could be the other number: 2, 3, 5, 8, 10?

Solution

‘Idli’ is said for multiples of the smaller number (here 4) and ‘vada’ for multiples of the larger number. Nobody said just ‘vada’, which means every multiple of the larger number is also a multiple of 4 — so it always becomes ‘idli-vada’.

This happens only when the larger number is itself a multiple of 4. Among 2, 3, 5, 8, 10 the only multiple of 4 is 8.

Check: multiples of 8 are \(8, 16, 24, 32,\dots\) — each one is also a multiple of 4, so ‘vada’ alone is never said.

The other number is 8

What jump size can reach both 15 and 30? Find all of them.

Solution

A jump size lands on a number only if it is a factor of that number. To land on both, the 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: \(1, 3, 5, 15\)

Jump sizes: 1, 3, 5, 15

Look at the table. What do you notice about the shaded and circled numbers?

Number grid 31 to 70 with multiples of 3 shaded and multiples of 4 circled — Numbers 31–70: multiples of 3 are shaded, multiples of 4 are circled.

Solution

1. Shaded numbers: the shaded numbers (33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69) are all multiples of 3.

2. Circled numbers (32, 36, 40, 44, 48, 52, 56, 60, 64, 68) are all multiples of 4.

3. Both shaded and circled: \(36, 48, 60\). These are the common multiples of 3 and 4 — every one is a multiple of \(3\times4=12\).

36, 48, 60 — common multiples of 3 and 4

Figure it Out page 108

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

Solution

‘Idli-vada’ is said at numbers that are multiples of both 3 and 5 — that is, the multiples of 15.

\(15, 30, 45, 60, 75, 90, 105, 120, 135, \mathbf{150}\)

The 10th multiple of 15 is \(15\times10\).

\(15\times 10 = 150\)

If the game is played for the numbers 1 to 90, find out:

How many times would the children say ‘idli’ (including ‘idli-vada’)?
How many times would they say ‘vada’ (including ‘idli-vada’)?
How many times would they say ‘idli-vada’?

Solution

a. ‘Idli’ = every multiple of 3 up to 90:

\(\left\lfloor \dfrac{90}{3}\right\rfloor = 30\)

b. ‘Vada’ = every multiple of 5 up to 90:

\(\left\lfloor \dfrac{90}{5}\right\rfloor = 18\)

c. ‘Idli-vada’ = every multiple of 15 up to 90:

\(\left\lfloor \dfrac{90}{15}\right\rfloor = 6\)

a) 30 times · b) 18 times · c) 6 times

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

Solution

Use the same idea, now up to 900:

‘Idli’ (multiples of 3): \(\dfrac{900}{3} = 300\)

‘Vada’ (multiples of 5): \(\dfrac{900}{5} = 180\)

‘Idli-vada’ (multiples of 15): \(\dfrac{900}{15} = 60\)

idli 300 · vada 180 · idli-vada 60

Is this figure somehow related to the ‘idli-vada’ game? Hint: imagine playing till 30, then draw the figure for the game played till 60.

Venn diagram of multiples of 3 and multiples of 5 with common multiples in the middle — Fig. 5.1 — Multiples of 3 and Multiples of 5.

Solution

Yes. The left circle holds the ‘idli’ numbers (multiples of 3), the right circle holds the ‘vada’ numbers (multiples of 5), and the overlap holds the numbers that are multiples of both — exactly the ‘idli-vada’ numbers (multiples of 15).

Playing the game up to 60, the diagram looks like this:

Venn diagram for multiples of 3 and 5 up to 60, with 15, 30, 45, 60 in the overlap — Game played till 60 — overlap = idli-vada numbers 15, 30, 45, 60.

Yes — the overlap = idli-vada numbers

Figure it Out pages 110–111

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

Solution

List multiples of 40 and keep those strictly between 310 and 410:

\(40\times7 = 280\;(\text{too small})\)

\(40\times8 = 320,\quad 40\times9 = 360,\quad 40\times10 = 400\)

\(40\times11 = 440\;(\text{too big})\)

320, 360, 400

Who am I?

I am less than 40. One of my factors is 7. The sum of my digits is 8.
I am 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: \(7, 14, 21, 28, 35\). Their digit sums are \(7, 5, 3, 10, 8\). Only \(35\) gives a digit sum of \(3+5=8\).

b. Factors 3 and 5 ⇒ a multiple of 15: \(15, 30, 45, 60, 75, 90\). We need digits differing by 1 → \(45\) (since \(5 = 4+1\)).

a) 35 · b) 45

A perfect number is one whose factors add up to twice the number (e.g. 28: factors 1, 2, 4, 7, 14, 28 sum to 56 = 2×28). Find a perfect number between 1 and 10.

Solution

Try \(6\). Its factors are \(1, 2, 3, 6\).

\(1 + 2 + 3 + 6 = 12 = 2\times 6\) ✓

6 is a perfect number

Find the common factors of:

20 and 28
35 and 50
4, 8 and 12
5, 15 and 25

Solution

a. \(20\!:\,1,2,4,5,10,20\) · \(28\!:\,1,2,4,7,14,28\) → common 124

b. \(35\!:\,1,5,7,35\) · \(50\!:\,1,2,5,10,25,50\) → common 15

c. \(4,8,12\) → common 124

d. \(5,15,25\) → common 15

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

Solution

Multiples of 50 are the even multiples of 25 (50, 100, 150…). So the odd multiples of 25 work:

\(25,\;75,\;125\quad(\text{also }175, 225,\dots)\)

25, 75, 125

Anshu plays the ‘idli-vada’ game with two numbers, both smaller than 10. The first ‘idli-vada’ is said after the number 50. What could the two numbers be?

Solution

The first ‘idli-vada’ happens at the first common multiple (the LCM) of the two numbers. We need two numbers below 10 whose first common multiple is more than 50.

\(7 \text{ and } 8:\ \text{LCM} = 56 > 50\) ✓

\(8 \text{ and } 9:\ \text{LCM} = 72 > 50\) ✓

7 & 8, or 8 & 9

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

Solution

The jump sizes are the common factors of 28 and 70.

\(28\!:\,1,2,4,7,14,28\)

\(70\!:\,1,2,5,7,10,14,35,70\)

Common: \(1, 2, 7, 14\)

Jump sizes: 1, 2, 7, 14

Guna erased all the numbers except the common multiples (24, 48, 72). Find what the two numbers could be and fill the empty regions.

Two overlapping circles with 72, 48 and 24 in the common region — Only the common multiples 24, 48, 72 are left.

Solution

The common multiples shown are \(24, 48, 72\) — these are consecutive multiples of \(24\). So the two numbers must have a first common multiple (LCM) of 24.

A neat choice is 3 and 8 (since \(\text{LCM}(3,8)=24\)). Then:

Multiples of 3 only: 3, 6, 9, 12, 15, 18, 21, 27, 30, …
Multiples of 8 only: 8, 16, 32, 40, 56, 64, …
Common (centre): 24, 48, 72

Other pairs also work, e.g. 6 and 8 or 8 and 24 — any pair whose LCM is 24.

e.g. Multiples of 3 & Multiples of 8

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

Solution

Take the highest power of each prime needed by 1–10 (leaving out 7):

\(2^3\) (from 8), \(3^2\) (from 9), \(5\) (from 5, 10)

\(2^3\times 3^2\times 5 = 8\times 9\times 5 = 360\)

360

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

Solution

Now we also include the prime 7:

\(2^3\times 3^2\times 5\times 7 = 360\times 7 = 2520\)

2520

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5.2

Prime Numbers

In-text Question page 113

How many prime numbers are there from 21 to 30? How many composite numbers?

Sieve of Eratosthenes grid from 1 to 100 with primes circled — The Sieve of Eratosthenes — primes circled, composites crossed out.

Solution

From 21 to 30 there are 10 numbers.

Primes: \(23, 29\) → 2 primes

Composites: \(21, 22, 24, 25, 26, 27, 28, 30\) → 8 composites

2 primes, 8 composites

Figure it Out page 114

2 is a prime and also even. Is there any other even prime?

Solution

No. Every even number other than 2 has 2 as a factor, so it has at least three factors (1, 2 and itself) — making it composite.

2 is the only even prime

Looking at the primes up to 100, what is the smallest difference between two successive primes? The largest?

Solution

Smallest: \(3 - 2 = 1\) (only between 2 and 3).

Largest: \(97 - 89 = 8\).

Smallest = 1 · Largest = 8

Are there an equal number of primes in every row (decade) up to 100? Which decades have the fewest primes? The most?

Solution

No — the count varies.

Most primes: 1–10 and 11–20 have 4 each (2, 3, 5, 7 and 11, 13, 17, 19).

Fewest primes: 91–100 has just 1 (only 97).

Most: 1–10 & 11–20 · Fewest: 91–100

Which of these are prime: 23, 51, 37, 26?

Solution

\(23\) → prime ✓

\(51 = 3\times 17\) → composite

\(37\) → prime ✓

\(26 = 2\times 13\) → composite

23 and 37 are prime

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

Solution

\(2 + 3 = 5\)

\(3 + 7 = 10\)

\(2 + 13 = 15\)

(2, 3), (3, 7), (2, 13)

Other pairs also work, such as \((7, 13)\) since \(7+13 = 20\).

13 and 31 are both prime and use the same digits reversed. Find such pairs up to 100.

Solution

Look for two-digit primes whose reversal is also prime:

\(17 \leftrightarrow 71,\quad 37 \leftrightarrow 73,\quad 79 \leftrightarrow 97\)

17 & 71, 37 & 73, 79 & 97

Find seven consecutive composite numbers between 1 and 100.

Solution

The stretch just before 97 has a long run with no primes:

\(90, 91, 92, 93, 94, 95, 96\)

(Here \(91 = 7\times 13\) — none of these seven is prime.)

90, 91, 92, 93, 94, 95, 96

Twin primes are primes that differ by 2 (like 3 & 5, or 17 & 19). Find the other twin primes between 1 and 100.

Solution

(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

8 twin-prime pairs up to 100

Identify whether each statement is true or false, and explain.

There is no prime number whose units digit is 4.
A product of primes can also be prime.
Prime numbers do not have any factors.
All even numbers are composite numbers.
2 is a prime and so is 3. For every other prime, the next number is composite.

Solution

a. True. A units digit of 4 makes the number even, so it is divisible by 2 and cannot be prime.

b. False. A product of two or more primes has those primes as factors, so it is composite (e.g. \(2\times3 = 6\)).

c. False. Primes have exactly two factors — 1 and the number itself.

d. False. 2 is even but prime; only even numbers greater than 2 are composite.

e. True. Any prime greater than 2 is odd, so the next number is even and greater than 2 — hence composite. Only 2 & 3 are consecutive primes.

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

Solution

\(45 = 3\times 3\times 5\) — only two distinct primes

\(60 = 2\times 2\times 3\times 5\) — a prime repeats

\(91 = 7\times 13\) — only two primes

\(105 = 3\times 5\times 7\) — exactly three distinct primes ✓

\(330 = 2\times 3\times 5\times 11\) — four distinct primes

\(105 = 3\times 5\times 7\)

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

Solution

The possible numbers are \(245, 254, 425, 452, 524, 542\).

Every one ends in \(2, 4\) or \(5\), so each is divisible by 2 or 5 — none can be prime.

None (zero)

3 is prime and \(2\times3+1 = 7\) is also prime. Find at least five other primes for which doubling and adding 1 gives another prime.

Solution

\(2\times 2+1 = 5\) ✓

\(2\times 5+1 = 11\) ✓

\(2\times 11+1 = 23\) ✓

\(2\times 23+1 = 47\) ✓

\(2\times 29+1 = 59\) ✓

e.g. 2, 5, 11, 23, 29

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5.3

Co-prime Numbers for Safekeeping Treasures

In-text Questions pages 115–116

Grumpy wants pairs that Jumpy cannot reach with any jump size except 1 (a “safe” pair). Check if these pairs are safe: (a) 15 & 39, (b) 4 & 15, (c) 18 & 29, (d) 20 & 55.

Solution

A pair is “safe” when the two numbers share no common factor other than 1.

a. \(15\) & \(39\) share the factor 3 → not safe.

b. \(4\) & \(15\): factors \(\{1,2,4\}\) and \(\{1,3,5,15\}\) → only 1 in common → safe.

c. \(18\) & \(29\): 29 is prime and doesn’t divide 18 → only 1 in common → safe.

d. \(20\) & \(55\) share the factor 5 → not safe.

Safe pairs: 4 & 15, and 18 & 29

Which of the following pairs are co-prime: (a) 18 & 35, (b) 15 & 37, (c) 30 & 415, (d) 17 & 69, (e) 81 & 18?

Solution

Two numbers are co-prime if their only common factor is 1.

a. \(18 = 2\times 3^2,\ 35 = 5\times 7\) → no common prime → co-prime ✓

b. \(15 = 3\times 5,\ 37\) is prime → co-prime ✓

c. \(30 = 2\times 3\times 5,\ 415 = 5\times 83\) → share 5 → not co-prime

d. \(17\) prime, \(69 = 3\times 23\) → co-prime ✓

e. \(81 = 3^4,\ 18 = 2\times 3^2\) → share 3 → not co-prime

Co-prime: a, b and d

While playing ‘idli-vada’ with different pairs, Anshu noticed: sometimes the first common multiple equals the product of the two numbers, and sometimes it is less. Find examples of each. How is this related to being co-prime?

Solution

First common multiple = product (the numbers are co-prime):

\((3,5)\to 15 = 3\times 5\); \((3,7)\to 21\); \((4,9)\to 36\)

First common multiple < product (they share a factor):

\((3,6)\to 6 < 18\); \((3,12)\to 12 < 36\); \((6,15)\to 30 < 90\)

Relation: the first common multiple equals the product exactly when the two numbers are co-prime. If they share a factor, the first common multiple is smaller.

Product ⇔ the pair is co-prime

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5.4

Prime Factorisation

Every composite number can be broken down into a product of primes — its prime factorisation. A factor tree makes this easy:

Four different factor trees of 36, all ending in two 2s and two 3s — Four ways to factorise 36 — each ends with \(2\times2\times3\times3\).

Figure it Out page 120

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

Solution

\(64 = 2^6\)

\(104 = 2^3\times 13\)

\(105 = 3\times 5\times 7\)

\(243 = 3^5\)

\(320 = 2^6\times 5\)

\(141 = 3\times 47\)

\(1728 = 2^6\times 3^3\)

\(729 = 3^6\)

\(1024 = 2^{10}\)

\(1331 = 11^3\)

\(1000 = 2^3\times 5^3\)

A number’s prime factorisation has one 2, two 3s, and one 11. What is the number?

Solution

\(2\times 3\times 3\times 11 = 2\times 9\times 11\)

\(= 198\)

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

Solution

1955 ends in 5, so divide by 5:

\(1955 = 5\times 391\)

\(391 = 17\times 23\)

\(1955 = 5\times 17\times 23\)

Find the prime factorisation of these without multiplying first:

56 × 25
108 × 75
1000 × 81

Solution

Factorise each part, then combine:

a. \(56\times 25 = (2^3\times 7)\times(5^2) = 2^3\times 5^2\times 7\)

b. \(108\times 75 = (2^2\times 3^3)\times(3\times 5^2) = 2^2\times 3^4\times 5^2\)

c. \(1000\times 81 = (2^3\times 5^3)\times(3^4) = 2^3\times 3^4\times 5^3\)

What is the smallest number whose prime factorisation has (a) three different primes? (b) four different primes?

Solution

Use the smallest primes, once each:

a. \(2\times 3\times 5 = 30\)

b. \(2\times 3\times 5\times 7 = 210\)

a) 30 · b) 210

Figure it Out page 122

Are these pairs co-prime? Guess, then verify with prime factorisation: (a) 30 & 45, (b) 57 & 85, (c) 121 & 1331, (d) 343 & 216.

Solution

a. No. \(30 = 2\times 3\times 5,\ 45 = 3^2\times 5\) → share 3 and 5.

b. Yes. \(57 = 3\times 19,\ 85 = 5\times 17\) → no common prime.

c. No. \(121 = 11^2,\ 1331 = 11^3\) → share 11.

d. Yes. \(343 = 7^3,\ 216 = 2^3\times 3^3\) → no common prime.

Co-prime: b and d

Is the first number divisible by the second? Use prime factorisation: (a) 225 & 27, (b) 96 & 24, (c) 343 & 17, (d) 999 & 99.

Solution

The second divides the first only if its whole prime factorisation fits inside the first’s.

a. No. \(225 = 3^2\times 5^2,\ 27 = 3^3\). 27 needs three 3s but 225 has only two.

b. Yes. \(96 = 2^5\times 3,\ 24 = 2^3\times 3\) fits inside → \(96 = 24\times 4\).

c. No. \(343 = 7^3\); 17 is a prime that isn’t a factor.

d. No. \(999 = 3^3\times 37,\ 99 = 3^2\times 11\); 11 isn’t a factor of 999.

Divisible only in (b)

The first number is \(2\times3\times7\) and the second is \(3\times7\times11\). Are they co-prime? Does one divide the other?

Solution

They share the primes 3 and 7, so they are not co-prime.

Neither factorisation fits completely inside the other (the first has a 2 the second lacks; the second has an 11 the first lacks), so neither divides the other.

Not co-prime · neither divides the other

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

Solution

Yes. Two different primes have no factors except 1 and themselves, so their only common factor is 1.

e.g. \(2\) & \(3\), or \(3\) & \(11\) — always co-prime.

Yes — any two different primes are co-prime

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5.5

Divisibility Tests

In-text Questions pages 124–125

Is 8536 divisible by 4?

Solution

For divisibility by 4, only the last two digits matter.

Last two digits: \(36\), and \(36\div 4 = 9\) — exact.

Yes, 8536 is divisible by 4

Do you agree with these statements about divisibility by 4? (1) Only the last two digits matter. (2) If the last two digits form a number divisible by 4, so is the whole number. (3) If the whole number is divisible by 4, so is the number made by its last two digits.

Solution

All three are true. Every hundred (100, 200, 300…) is a multiple of 4, so the hundreds/thousands part never affects divisibility by 4 — only the last two digits decide it.

e.g. \(124, 364, 4028\) are all divisible by 4, matching their last two digits (24, 64, 28).

1 ✓ · 2 ✓ · 3 ✓

Find numbers between 120–140, 1120–1140, and 3120–3140 that are divisible by 8. What do you observe?

Solution

120–140: \(128, 136\)

1120–1140: \(1128, 1136\)

3120–3140: \(3128, 3136\)

Observation: the same endings (…128, …136) repeat — for divisibility by 8 only the last three digits matter.

128/136 pattern repeats

Change the last two digits of 8560 so the result is a multiple of 8.

Solution

Keep 85 and pick last two digits making the last three digits divisible by 8:

\(8552\) → last three digits \(552 = 8\times 69\) ✓

e.g. 8552

Figure it Out pages 125–126

Leap years are multiples of 4, except century years not divisible by 400. (a) From your birth year till now, which years were leap years? (b) From 2024 till 2099, how many leap years are there?

Solution

a. This depends on your birth year — list the multiples of 4 from then until now (e.g. 2012, 2016, 2020, 2024 …), skipping any century year not divisible by 400.

b. Leap years from 2024 to 2099 are the multiples of 4: \(2024, 2028, \dots, 2096\) (2100 is beyond the range).

Count \(= \dfrac{2096 - 2024}{4} + 1 = \dfrac{72}{4} + 1 = 18 + 1 = 19\)

b) 19 leap years

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

Solution

A 4-digit palindrome looks like \(\overline{abba}\). It is divisible by 4 only when its last two digits \(\overline{ba}\) are divisible by 4.

Largest: if \(a = 9\), the ending \(\overline{b9}\) is odd — never divisible by 4. Trying \(a = 8\) gives \(\overline{b8}\); the largest works at \(b = 8\):

\(8888\) → \(88 \div 4 = 22\) ✓

Smallest: \(a = 1\) gives an odd ending. With \(a = 2\), \(\overline{b2}\) is divisible by 4 at \(b = 1\):

\(2112\) → \(12 \div 4 = 3\) ✓

Largest 8888 · Smallest 2112

Is each statement always, sometimes, or never true? (a) Sum of two even numbers is a multiple of 4. (b) Sum of two odd numbers is a multiple of 4.

Solution

a. Sometimes true.

\(2 + 6 = 8\) ✓ but \(2 + 4 = 6\) ✗

b. Sometimes true.

\(1 + 3 = 4\) ✓ but \(1 + 5 = 6\) ✗

Both are sometimes true

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

Solution

The last digit alone gives all three remainders.

Number	÷ 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

The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked only two of these and declared it divisible by all. Which two numbers could they be?

Solution

Checking 8 automatically covers 2 and 4 (any multiple of 8 is a multiple of 2 and 4). Checking 5 together with divisibility by 2 covers 10.

Divisible by 8 ⇒ divisible by 2 and 4; divisible by 5 and 2 ⇒ divisible by 10.

5 and 8

Which of these are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160?

Solution

The number must end in 0 (for 10) and have its last three digits divisible by 8.

\(572\) — doesn’t end in 0 ✗

\(2352\) — doesn’t end in 0 ✗

\(5600\) — ends in 0, and \(600 \div 8 = 75\) ✓

\(6000\) — ends in 0, and \(000 \div 8 = 0\) ✓

\(77622160\) — ends in 0, and \(160 \div 8 = 20\) ✓

5600, 6000, 77622160

Write two numbers whose product is 10000, where neither number has 0 as its units digit.

Solution

Factorise: \(10000 = 2^4\times 5^4\). A trailing 0 appears only when a number has both a 2 and a 5. So put all the 2s in one number and all the 5s in the other: