Free access of the complete Ganita Prakash Book Class 6 Solutions and Chapter 5 Prime Time Class 6 NCERT Solutions Question Answer are crafted in simple format to align with the latest CBSE syllabus.

## Class 6 Maths Chapter 5 Prime Time Solutions

### Prime Time Class 6 Solutions Questions and Answers

**5.1 Common Multiples and Common Factors Figure it Out (Page No. 108)**

Question 1.

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

Solution:

The first number for which the players should say, ‘idli-vada’ is 15 as 3 × 5, which is a multiple of 3 and 5.

So, 15 × 1 = 15,

15 × 2 = 30,

15 × 3 = 45,

………………….

………………….

………………….

15 × 10 = 150

Thus, 150 is the number that will be said ‘idli-vada’ for the 10th time.

Question 2.

If the game is played for the numbers from 1 till 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 from number 1 till 90: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90

Total number of multiples of 3 from number 1 to 90 = 30

Thus, the children would say 30 times ‘idli’ (including the times they say ‘idli-vada’).

(b) Multiples of 5 from number 1 till 90: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.

Total number of multiples of 5 from number 1 to 90 = 18

Thus, the children would say 18 times ‘vada’ (including the times they say ‘idli-vada’).

(c) Numbers that are multiples of 3 and 5 both from 1 to 90 are: 15, 30, 45, 60, 75, and 90.

Total number of multiples of 15 from number 1 to 90 = 6.

Thus, the children would say ‘idli-vada’ 6 times.

Question 3.

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

Solution:

If the game was played till 900 that is 90 × 10.

Then, total number of multiples of 3 from number 1 to 900 = 30 × 10 = 300

Thus, the children would say 300 times ‘idli’ (including the times they say ‘idli-vada’).

Similarly, total number of multiples of 5 from number 1 to 900= 18 × 10= 180

Thus, the children would say 180 times ‘vada’ (including the times they say ‘idli-vada’).

And, total number of multiples of 15 from number 1 to 900 = 6 × 10 = 60

Thus, the children would say 60 times ‘idli-vada’.

Question 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 given figure is related to the ‘idli-vada’ game. The numbers for ‘idli’ are given in the left circle, the numbers for ‘vada’ are given in the right circle and the numbers common in both the circles are for’idli-vada’.

Intext Questions

Question 1.

Let us now play the ‘idli-vada’ game with different pairs of numbers: (Page 108)

(a) 2 and 5,

(b) 3 and 7,

(c) 4 and 6.

We will say ‘idli’ for multiples … game is played up to 60.

Solution:

Question 2.

Which of the following could be the other number:

2,3, 5, 8, 10? (Page 109)

Solution:

As one of the numbers was 4. Among the given numbers, every multiple of 8 is also a multiple of 4. So, the other number is 8.

Question 3.

What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. (Page 110)

Solution:

Factors of 15 are 1, 3, 5 and 15

Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30

The jump size that reach both 15 and 30 are 1, 3, 5 and 15, the common factors of 15 and 30.

Question 4.

Look at the table below. What do you notice?

In the table,

1. Is there anything common among the shaded numbers?

Solution:

The shaded numbers are the multiples of 3 starting from 33.

2. Is there anything common among the circled numbers?

Solution:

The circled numbers are the multiples of 4 starting from 32.

3. Which numbers are both shaded and circled? What are these numbers called?

Solution:

Numbers 36,48 and 60 are both shaded and circled. These numbers are called common multiples of 3 and 4.

**5.1 Common Multiples and Common Factors Figure it Out (Page No. 110 – 111)**

Question 1.

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

Solution:

All multiples of 40 that lie between 310 and 410:

40 × 8 = 320, 40 × 9 = 360, and 40 × 10 = 400

Thus, all multiples of 40 that lie between 310 and 410 are 320, 360 and 400.

Question 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) Numbers that is less than 40 and have 7 as factor:

7 × 1 = 7,

7 × 2 = 14,

7 × 3 = 21,

7 × 4 = 28,

7 × 5 = 35

Here, only 35 is the number whose sum of the digits is 8 as 3 + 5 = 8.

Hence, the number is 35.

(b) Numbers less than 100 having two of their factors 3 and 5 are:

1 × (3 × 5) = 15, 2 × (3 × 5) = 30, 3 × (3 × 5) = 45,

4 × (3 × 5) = 60, 5 × (3 × 5) = 75, 6 × (3 × 5) = 90

Here, 45 is the only number whose one of the digits is 1 more than the other (4 + 1 = 5).

Question 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:

There is only one number 6 between 1 to 10 which is a perfect number.

As, the factors of 6 are 1, 2, 3 and 6 and their sum is 1 + 2 + 3 + 6 = 12, which is twice the number 6.

Question 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) 20 and 28.

We have: 20 = 1 × 20, 20 = 2 × 10, 20 = 4 × 5

∴ All the factors of 20 are: 1, 2, 4, 5, 10 and 20 …(i)

Again, 28 = 1 × 28, 28 = 2 × 14, 28 = 4 × 7

∴ All the factors of 28 are: 1, 2, 4, 7, 14 and 28 …(ii)

From (i) and (ii), common factors of 20 and 28 are 1, 2 and 4.

(b) 35 and 50

Since 35 = 1 × 35, 35 = 5 × 7

All the factors of 35 are: 1, 5, 7 and 35 …(i)

Again, 50 = 1 × 50, 50 = 2 × 25, 50 = 5 × 10

All the factors of 50 are: 1, 2, 5, 10, 25, 50 …(ii)

From (i) and (ii), common factors of 35 and 50 are 1 and 5.

(c) 4, 8 and 12

∵ Factors of 4 are: 1,2 and 4 Factors of 8 are: 1, 2, 4 and 8

Factors of 12 are: 1, 2, 3, 4, 6 and 12

Common factors of 4, 8 and 12 are 1, 2 and 4.

(d) 5, 15 and 25

∵ Factors of 5 are: 1 and 5 Factors of 15 are: 1, 3, 5 and 15

Factors of 25 are: 1, 5 and 25

∴ Common factors of 5, 15 and 25 are 1 and 5.

Question 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, 175, …

Multiples of 50 are: 50, 100, 150, 200, 250, …

∴ The numbers that are multiples of 25 but not multiples of 50 are 25, 75, 125, 175,…

Question 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 pair of numbers smaller than 10 with first common multiple greater than 50 are (7, 8), (8, 9) and (7, 9).

For (7, 8):

If 7, 14, 21, 28, 35, …, 49, 56, 63, … are said ‘idli’.

And 8, 16, 24, 32, …, 48, 56, 64, … are said ‘vada’.

Then 56 is the first number (first common multiple of 7 and 8) greater than 50 which would be said ‘idli-vada’.

For (8, 9):

If 8, 16, 24, 32, …, 48, 56, 64, 72, … are said ‘idli’.

And 9, 18, 27, 36, …, 54, 63, 72, … are said ‘vada’.

Then 72 is the first number (first common multiple of 8 and 9) greater than 50 which would be said ‘idli-vada’.

For (7, 9):

If 7, 14, 21, 28, …, 49, 56, 63, 70, … are said ‘idli’.

And 9, 18, 27, 36, …, 54, 63, 72, … are said ‘vada’.

Then 63 is the first number (first common multiple of 7 and 9) greater than 50 which would be said ‘idli-vada’.

Question 7.

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

Solution:

In the treasure hunting game, if Grumpy has kept treasures on 28 and 70, then the common factor of 28 and 70 will be the jump sizes to land on both the numbers.

Therefore,

Factors of 28 are: 1, 2, 4, 7, 14, and 28.

Factors of 70 are: 1, 2, 5, 7, 10, 14, 35 and 70.

The common factors of 28 and 70 are: 1, 2, 7, and 14. Thus, the possible jump sizes to land on both the numbers will be 1, 2, 7, and 14.

Question 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.

Solution:

A pair of possible number

whose common multiples are 24, 48 and 72 would be 6 and 8.

Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,…

Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, …

Other possible pairs are 3 and 8; 8 and 12.

Question 9.

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

Solution:

The smallest number that is a multiple of all the numbers from 1 to 10 except for 7 will be the least common multiple of these numbers.

2 × 2 × 3 × 5 × 2 × 3 = 360

Thus, the smallest number that is a multiple of all the numbers from 1 to 10 except for 7 will be 360.

Question 10.

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

Solution:

The smallest number that is a multiple of all the numbers from 1 to 10 will be least common multiples of that numbers.

2 × 2 × 3 × 5 × 7 × 2 × 3 = 2520

Thus, the smallest number that is a multiple of all the numbers from 1 to 10 will be 2520.

InText Questions

Question 1.

How many prime numbers are there from 21 to 30? How many composite numbers are there from 21 to 30? (Page 113)

Solution:

There are 2 prime numbers, (i.e., 23,29) and 8 composite numbers (i.e., 21,22, 24, 25, 26, 27, 28 and 30) from 21 to 30.

**5.2 Prime Numbers Figure it Out (Page No. 114 – 115)**

Question 1.

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

Solution:

No, 2 is the only prime which is even also.

Question 2.

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

Solution:

Looking at the list of primes till 100, we have

The smallest difference between two successive primes is 1 as 3 – 2 = 1.

And the largest difference between two successive primes is 8 as 97 – 89 = 8.

Question 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, from 91 to 100, there is only 1 prime number which is 97.

From 1 to 10 and 11 to 20, there are 4 prime numbers in each of the two rows.

Question 4.

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

Solution:

Among the given numbers 23 and 37 are prime numbers as they have only two factors 1 and the numbers themselves.

Question 5.

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

Solution:

The pairs of prime numbers less than 20 whose sum is a multiple of 5 are as follows:

(2, 3) as 2 + 3 = 5 which is a multiple of 5.

(3, 7) as 3 + 7 = 10 which is a multiple of 5.

(2, 13) as 2 + 13 = 15 which is a multiple of 5.

(7, 13) as 7 + 13 = 20 which is a multiple of 5.

(3, 17) as 3 + 17 = 20 which is a multiple of 5.

(Answer may vary)

Question 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:

Required pairs of prime numbers up to 100 having the same digits are: 17 and 71; 37 and 73; 79 and 97.

Question 7.

Find seven consecutive composite numbers between 1 and 100.

Solution:

Seven consecutive composite numbers between 1 and 100 are: 90, 91, 92, 93, 94, 95, and 96.

Question 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:

The twin primes (other than 3 and 5; 17 and 19) between 1 and 100: (5, 7), (11, 13), (29, 31), (41, 43), (59, 61), and (71, 73).

Question 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 number ending with 4 is an even number and every even number is a multiple of 2. So, no prime number is possible with units digit 4.

(b) False. The product of two primes can’t be a prime because it violates the definition of the prime number as it will be divided by the prime numbers which are multiplied rather than 1 and the number itself. So the number is not prime it will become a composite number instead.

(c) False. Prime numbers have only two factors: 1 and the number itself.

(d) False. 2 is the smallest even prime number.

(e) True. 2 is the only even prime number, and 3 is the next prime after 2. For all other primes greater than 3, the next number is composite. For example, after 5 (which is prime), the next number is 6 (composite), and after 7 (prime), the next number is 8 (composite).

Question 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. They are not distinct as it has two 3’s.

60 = 2 × 2 × 3 × 5. They are not distinct as it has two 2’s.

91 = 7 × 13. It has only two distinct prime factors.

105 = 3 × 5 × 7. It has three distinct prime factors 3, 5, and 7.

330 = 2 × 3 × 5 × 11. It has four distinct prime factors.

Thus, only 105 is the product of 3 distinct prime numbers.

Question 11.

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

Solution:

The possible numbers formed by 2,4 and 5 are: 245, 254, 425, 452, 524, 542.

These all are composite numbers not prime numbers. So, no prime number can be made using each of 2,4 and 5 once.

Question 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:

We can have primes as:

2 × 2 + 1 = 5, prime number;

2 × 5 + 1 = 11, prime number;

2 × 11 + 1= 23, prime number;

2 × 23 + 1 = 47, prime number;

2 × 29 + 1 = 59, primfe number;

and 2 × 41 + 1 = 83, prime number;

Thus, 2, 5, 11, 23, 29 and 41 are required primes. (Answer may vary)

InText Questions

Question 1.

Which of the following pairs of numbers are co-prime? (Page 116)

(a) 18 and 35

(b) 15 and 37

(c) 30 and 415

(d) 17 and 69

(e) 81 and 18

Solution:

(a) 18 and 35

Factors of 18 are 1, 2, 3, 6, 9 and 18

Factors of 35 are: 1, 5, 7 and 35

Since, 18 and 35 have no common factor except 1.

Thus, 18 and 35 are co-prime numbers.

(b) 15 and 37

Factors of 15 are: 1, 3, 5 and 15

Factors of 37 are: 1, 37

Since, 15 and 37 have no common factor except 1.

Thus, 15 and 37 are co-prime numbers.

(c) 30 and 415

Factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30

Factors of 415 are: 1,5,83 and 415

Their common factors are: 1 and 5

Thus, 30 and 415 are not co-prime numbers.

(d) 17 and 69

Factors of 17 are: 1 and 17

Factors of 69 are: 1, 3, 23 and 69

Since, 17 and 69 have no common factor except 1.

Thus, 17 and 69 are co-prime numbers.

(e) 81 and 18

Factors of 81 are: 1, 3, 9, 27 and 81

Factors of 18 are: 1, 2, 3, 6, 9 and 18

Common factors of 81 and 18 are 3 and 9 other than 1.

Thus, 81 and 18 are not co-prime numbers.

Question 2.

Make such pictures for the following: (Page 117)

(a) 15 pegs, thread-gap of 10

Solution:

(b) 10 pegs, thread-gap of 7

Solution:

(c) 14 pegs, thread-gap of 6

Solution:

(d) 8 pegs, thread-gap of 3

Solution:

**5.3 Co-prime Numbers for Safekeeping Treasures Figure it Out (Page No. 120)**

Question 1.

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

Solution:

Question 2.

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

Solution:

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

Therefore, required number = 2 × 3 × 3 × 11 = 198

Question 3.

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

Solution:

5 × 17 × 23 = 1955.

Thus, 5, 17 and 23 are three prime numbers 17 less than 30 whose product is 1955.

Question 4.

Find the prime factorisation of these numbers without multiplying first

(a) 56 × 25

(b) 108 × 75

(c) 1000 × 81

Solution:

We can find the prime factorisation of these numbers without multiplying first as follows:

(a) 56 × 25 = (2 × 2 × 2 × 7) × (5 × 5) = 2 × 2 × 2 × 5 × 5 × 7

(b) 108 × 75 = (2 × 2 × 3 × 3 × 3) × (3 × 5 × 5) = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

(c) 1000 × 81 =(2 × 2 × 2 × 5 × 5 × 5) × (3 × 3 × 3 × 3) = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Question 5.

What is the smallest number whose prime factorisation has:

(a) three different prime numbers?

(b) four different prime numbers?

Solution:

(a) The smallest number whose prime factorisation has three different prime numbers = 2 ×3 × 5 = 30

(b) The smallest number whose prime factorisation has four different prime numbers = 2 × 3 × 5 × 7 = 210

**5.4 Prime Factorisation Figure it Out (Page No. 122)**

Question 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 and 45

30 = 2 × 3 × 5

45 = 3 × 3 × 5

Clearly, 3 and 5 are the common factors of 30 and 45.

So, these are not co-prime numbers.

(b) 57 and 85

57 = 3 × 19

85 = 5 × 17

Clearly, there is no any common prime factor of 57 and 85.

So, these are co-prime numbers.

(c) 121 and 1331

121 = 11 × n

1331 = 11 × 11 × 11

Clearly, 11 and 11 are the common factors of 121 and 1331.

So, these are not co-prime numbers.

(d) 343 and 216

343 = 7 × 7 × 7

216 = 2 × 2 × 2 × 3 × 3 × 3

Clearly, there is no any common prime factor of 343 and 216.

So, these are co-prime numbers.

Question 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 and 27′

225 = 3 × 3 × 5 × 5

27 = 3 × 3 × 3

All prime factors of 27 are not included in the prime factors of 225.

This is because 3 occurs thrice in the prime factorisation of 27 but only twice in the prime factorisation of 225.

This means 225 is not divisible by 27.

(b) 96 and 24

96 = 2 × 2 × 2 × 2 × 2 × 3

24 = 2 × 2 × 2 × 3

All prime factors of 24 are also prime factors of 96, that is 24 (= 2 × 2 × 2 × 3) is included in the prime factorisation of 96.

This means 96 is divisible by 24.

(c) 343 and 17

343 = 7 × 7 × 7

17 = 17

Clearly, there is no common prime factor of 343 and 17.

So, these are co-prime numbers.

Hence, 343 is not divisible by 17.

(d) 999 and 99

999 = 3 × 3 × 3 × 37

99 = 3 × 3 × 11 ;

Only prime factors 3 × 3 of 99 is the prime factors of 999.

But the prime factor 11 of 99 is not included in the prime factorisation of 999.

This means 999 is not divisible by 99.

Question 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 first number = 2 × 3 × 7 = 42

The second = 3 × 7 × 11 = 231 (and 231 > 42)

Clearly, 3 is the common prime factor in both numbers. So, these numbers are not co-prime numbers.

And, only prime factors 3 × 7 of first number is the prime factors of second number. But the prime factor 2 of first number is not included in the prime factorisation of second number.

So, no one divides the other.

Question 4.

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

Solution:

Yes, he is right.

Any two prime numbers are co-prime to each other, as every prime number has only two factors 1 and the number itself, the only common factor of two prime numbers will be 1.

**5.5 Divisibility Tests Figure it Out (Page No. 15-17)**

Question 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 bom till now, which years were leap years?

(b) From the year 2024 till 2099, how many leap years are there?

Solution:

(a) Do it yourself.

(b) We know that leap years occur in the years that are multiples of 4, except century years.

Therefore, multiples of 4 after 2024, will be 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, and 2096, till 2099.

Therefore, there will be 19 leap years from the year 2024 till 2099.

Question 2.

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

Solution:

A palindromic number is a number that reads the same forward and backward.

The smallest 4-digit palindrome divisible by 4 is 2112. As, the number formed by last 2 digits of the number is divisible by 4 (12 – 4 = 3).

The greatest 4-digit palindrome which is divisible by 4 is 8888. As, the number formed by last 2-digit of the number is divisible by 4 (88 = 4 = 22).

Question 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. The sum of two even numbers is not always a multiple of 4. As,

2 (even) + 4 (even) = 6 (even) which is not multiple of 4.

2 (even) + 6 (even) = 8 (even) multiple of 4.

4 (even) + 6 (even) =10 (even) which is not multiple of 4.

4 (even) + 8 (even) = 12 (even) which is multiple of 4.

(b) Sometimes true. The sum of two odd numbers is not always a multiple of 4.

3 (odd) + 5 (odd) = 8 (even) which is a multiple of 4.

5 (odd) + 7 (odd) =12 (even) which is a multiple of 4.

1 (odd) + 5 (odd) = 6 (even) which is not divisible by 4. So, it is not a multiple of 4.

7 (odd) + 11 (odd) = 18 (even) which is not divisible by 4. So, it is not a multiple of 4.

Question 4.

Find the remainders obtained when each of the following numbers are divided by

(i) 10, (ii) 5, (iii) 2.

78, 99, 173, 572, 980, 1111,2345

Solution:

(i) When divided by 10:

78 ÷ 10: Q = 7 and R = 8;

99 ÷ 10: Q = 9 and R = 9;

173÷ 10: Q = 17 and R = 3;

572 ÷ 10: Q = 57 and R = 2;

980 ÷ 10: Q = 98 andR = 0;

1111 ÷ 10: Q = 111 and R = 1;

2345 ÷ 10: Q = 234 and R = 5

Note: To find the remainder when dividing a number by 10, simply check how much the last digit exceeds ‘O’.

(ii) When divided by 5

78 ÷ 5: Q= 15 andR = 3;

99 ÷ 5: Q = 19 and R = 4;

173 ÷ 5: Q = 34 andR = 3;

572 ÷ 5: Q = 114 and R = 2;

980 ÷ 5: Q= 196 and R = 0;

1111 ÷ 5: Q = 222 and R= 1;

2345 ÷ 5: Q = 469 and R = 0

Note: To find the remainder when dividing by 5, check how much the last digit exceeds the nearest ‘0’ or ‘5’.

(iii) When divided by 2

78 ÷ 2: Q = 39 and R = 0;

99 ÷ 2: Q = 49 and R = 1;

173 ÷ 2; Q = 86 and R = 1;

572 ÷ 2; Q = 286 and R = 0;

980 ÷ 2; Q = 490 and R = 0;

1111 ÷ 2: Q = 555 and R = 1;

2345 ÷ 2: Q= 1172 and R = 1.

Note: To find the remainder when dividing by 2, check if the given digits is even (remainder ‘0’) or odd (remainder ‘1’)-

Question 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:

Since, 14560 is an even number, and the number formed by last 3 digits (560) is divisible by 8 as 560 ÷ 8 = 70.

Thus, 14560 is divisible by 8.

Also, 14560 ends with 0 so, it is also divisible by 5 and 10 as well.

If a number is divisible by 8 and 10, it must be divisible by 2, 4 and 5 also.

So, Guna could have checked for divisibility by 8 and 10 to determine the number is divisible by all 2, 4, 5, 8 and 10.

Question 6.

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

Solution:

Question 7.

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

Solution:

Prime factorisation of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 = 16 × 625

Thus, 16 and 625 are two numbers that do not have 0 as the units digits but their product is 10000.

Intext Questions

Question 1.

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. (Pages 127-128)

Solution: