NCERT Class 6TH Maths Chapter 5: prime Time

Mathematics with solution May 07, 2025

In chapter 5 : Prime Time you will know about multiples, factors, prime numbers, composite numbers, co-prime numbers, pair of twin prime, divisibility rules and prime factorisation of numbers.

NCERT Class 6TH Maths Chapter 5:  prime Time




Multiples

When a number(first number) divide any other number(second number) with remainder zero then second number is multiple of first number.
(In simple language: All the numbers which appear in the table of given number are its multiples.)

Example:6, 12, 18, 24, ..... divisible by 6. All the numbers are multiple of 6.

Factors

Factors of any given number are those numbers by which the given number is exactly divided.

Example: 20 is divisible by 1, 2, 4, 5, 10, 20 . Therefore 1, 2, 4, 5, 10, 20 are factors of 20.

Topic 5.1 Common Multiples and Common Factors

Figure it Out

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

Solution:
On 150  'idli vada' said for 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:
First we write 1 to 90 counting then make  circle on multiple of 3 and cross on multiple of 5. 

(a)
Childern will say 'idli' =  total number of multiples of 3 from 1 to 90.
Total number of multiples of 3 = number of encircled number = 30
Hence, childern will say idli upto 30 times.

(b)
Childern will say 'vada' =  total number of multiples of 5 from 1 to 90.
Total number of multiples of 5 = number of crossed number = 18
Hence, childern will say idli upto 18 times.

(c)
Childern will say 'idli - vada' =  total number common of multiples of 3 and 5 from 1 to 90.
Total number common of multiples of 3 and 5 = number of encircled and crossed number = 6
Hence, children will say 'idli-vada' upto 6 times.

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

Solution:
If game will play till 900 then 'idli' will say upto 300 times, 'vada' will say upto 180 times and 'idli vada' will say upto 60 time. Each answer become by 10 times of previous (question1) answer.

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:

Question
Let us now play the ‘idli-vada’ game with different pairs of numbers:
a. 2 and 5,
b. 3 and 7,
c. 4 and 6.

Solution:
(a)

(b)
(c)

(Note: You can write more mutiples to find more common multiples)

Question
Which of the following could be the other number: 2, 3, 5, 8, 10?

Solution:
Here, 2 + 3 = 5, 3 + 5 = 8 and 5 + 8 = 13 so 10 is other number.

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

Solution:
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 size of 1, 3, 5, 15(multiple jumps) can reach both 15 and 30.

Question
Look at the table below. What do you notice?

In the 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?

Solution:
(1)
All the shaded numbers are multiples of 3.
(2)
All the encircled numbers are multiples of 4.
(3)
All encircled number with shaded are common multiples of 3 and 4.

Figure it Out

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

Solution:
Multiples of 40: 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, ....
Hence, all the multiples of 40 between 310 and 410 are 320, 360, 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) 
Number less than 40 that have a factor  7 = 7, 14, 21, 28, 35.
∵ 35 = 3 + 5 = 8
∴ Correct number is 35.

(b)
Number less than 100 that have  factors   3 and 5 = 15, 30, 45, 60, 75, 90.
In 45, 5 is one more than 4.
Hence, correct number is 45 

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:

Perfect number between 1 to 10 is 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) 
Factors of 20 = 1, 2, 4, 5, 10, 20.
Factors of 28 = 1, 2, 4, 7, 14, 28.
Common factors of 20 and 28 = 1, 2, 4.

(b)
Factors of 35 = 1, 5, 7, 35.
Factors of 50 = 1, 2, 5, 10, 25, 50.
Common factors of 20 and 28 = 1, 5

(c)
Factors of 4 = 1, 2, 4..
Factors of 8 = 1, 2, 4, 8.
Factors of 12 = 1, 2, 3, 4, 6, 12.
Common factors of 4, 8 and 12 = 1, 2, 4.
.
(d)
Factors of 5 = 1, 5.
Factors of 15 = 1, 3, 5, 15.
Factors of 25 = 1, 5, 25.
Common factors of 5, 15 and 25 = 1, 5.

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

Solution:
Multiples of 25 = 25, 50, 75, 100, 125, 150,.........
Multiples of 50 = 50, 100, 150, 200,....
Common multiples = 50, 100, 150,.....
Hence three numbers that are multiples of 25 but not multiples of 50 are 25, 75, 125 .

(Note: If you exclude the common multiples of 25 and 50 from mutiples of 25 then only multiples of 25 willremain)

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 ‘idlivada’ is after the number 50. What could the two numbers be
which are assigned ‘idli’ and ‘vada’?

Solution:
Pair of numbers less than 10 whose first common multiple is greater than 50 are  (7,8), (7,9) and (8,9).

First pair (7,8)
If we assigned  number 7 as 'idli' and number 8 as 'vada' then first time 'idli-vada' will say on 7✖8=56.


Second pair (7,9)
If we assigned  number 7 as 'idli' and number 9 as 'vada' then first time 'idli-vada' will say on 7✖9=63.


Third pair (8,9)
If we assigned  number 8 as 'idli' and number 9 as 'vada' then first time 'idli-vada' will say on 8✖9=72.

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:
Factors of 28 = 1, 2, 4, 7, 14, 28.
Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70.
Common factors = 1, 2, 7, 14.
Hence, the jump size may be any of 1, 2, 7, 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:
Possible pair whose common multiples are 24, 48 , 72 is  (6,8) or (3,8) or (12,24) or (8,24) or (8,12).
Now, we choose pair of (6,8)
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,.......
Common multiples = 24, 48, 72, .....


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 multiple of 1,2,3,4,5,6,8,9 = LCM of 1,2,3,4,5,6,8,9

 LCM of 1,2,3,4,5,6,8,9 = 2 x 2 x 2 x 3 x 3 x 5 = 360
Hence,The smallest number multiple of 1,2,3,4,5,6,8,9 = 360

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

Solution:
The smallest number multiple of 1,2,3,4,5,6,7,8,9 = LCM of 1,2,3,4,5,6,7,8,9

LCM of 1,2,3,4,5,6,8,9 = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520
Hence,The smallest number multiple of 1,2,3,4,5,6,7,8,9 = 2520

Topic 5.2 Prime Numbers

Numbers have only two factors 1 and number itself are called prime numbers.
(In simple word: Numbers that appear only their own table (Exception 1) are called prime numbers)

Example: 2, 3, 5, 7, 11, 13, ....... are prime numbers.

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

Solution:
There are only two prime numbers 23 and 29 from 21 to 30.
There are eight composite number 21, 22, 24, 25, 26, 27, 28 and 30 from 21 to 30.

Figure it Out

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

Solution:
Only 2 is even prime number there is no any other even prime.

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:
There is nothing between prime numbers 2 and 3.
then difference = 3 - 2 = 1

There are 7 numbers between prime numbers 89 and 97.
then difference = 97 - 89 = 8

Hence , the smallest difference is 1 and the largest difference is 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, there are not an equal number of primes occurring in every row in the table

There are only 1 prime number between 91 and 100.


From 1 to 10 and from 11 to 2o both decades have 4 prime numbers.

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

Solution:
Factors of 23 = 1, 23.
Factors of 51 = 1, 3, 17, 51.
Factors of 37= 1, 37
Factors of 26 = 1, 2, 13, 26.

Hence, 23 and 37 are prime numbers, 51 and 26 are composite numbers.

(Note:umbers have only two factors 1 and number itself are called prime numbers. If factyors are more than 2 are called composite numbers. )

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

Solution:
Prime numbers less than 20 = 2, 3, 5, 7, 11, 13, 19.
Now three pairs of prime numbers whose sum is multiple  of 5 are (2,3), (3,7) and (7,13).
2 + 3 = 5
3 + 7 = 10
7 + 13 = 20
5, 10, 20 are multiples of 5.

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:
Pairs of prime numbers with 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 are 90, 91, 92, 93, 94, 95, 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:
Pairs of twin primes between 1 and 100 are (3,5), (11,13), (17,19), (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, if unit digit of any number is 4 then it is composite number.
(b) False ,the prime numbers that are multiplied will become the factors of the product.
(c) False, Prime numbers have only 2 factors 1 and number itself.
(d) False, 2 is even prime numbers.
(e) True, only after 2 we get again prime number but for other prime number next number is composite. 

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

Solution:
First we find prime factorisation of each number
45 = 3 x 3 x 5
91 = 7 x 13
105 = 3 x 5 x 7
330 = 2 x 3 x 5 x 11

Hence ,105 is the product of exactly three distinct prime numbers.

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

Solution:
From the digits 2, 4 and 5 we can not make any prime number of 3 digits because all number wll form by these digits are composite. 

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:
2: 2 × 2 + 1 = 5 (prime)
3: 2 × 3 + 1 = 7(prime)
5: 2 × 5 + 1 = 11 (prime)
7: 2 × 7 + 1 = 15 (composite)
11: 2 × 11 + 1 = 23 (prime)
13: 2 × 13 + 1 = 27 (composite)
17: 2 × 17 + 1 = 35 (composite)
19: 2 × 19 + 1 = 39 (composite)
23: 2 × 23 + 1 = 47 (prime)
29: 2 × 29 + 1 = 59 (prime)
Hence, required prime numbers are 2, 3, 5, 11, 23, 29,.....

Topic 5.3 Co-prime numbers for safekeeping treasures

Two numbers are co-prime if they have only 1 as a common factor.

Example: (9 and 14), (15 and 16), (20 and 27) are pair of co prime numbers.
 

Which pairs are safe?

Let us go back to the treasure finding game. This time, treasures are kept on two numbers. Jumpy gets the treasures only if he is able to reach both the numbers with the same jump size. There is also a new rule — a jump size of 1 is not allowed. 

Question
Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures?Check if these pairs are safe:

a. 15 and 39             b. 4 and 15

 c. 18 and 29            d. 20 and 55

Solution:
(a)
Factors of 15 =1, 3, 5, 15
Factors of 39 = 1, 3, 13, 39
Common factors of 15 and 39 = 1, 3
By jump size 3 reach on both treasures.

(b)
Factors of 4 =1, 2, 4
Factors of 15 = 1, 3, 5, 15
Common factors of 4 and 15 = 1
Can not reach on both treasures because jump size of 1 not allowed.

(c)
Factors of 18 =1, 2, 3, 6, 9, 18
Factors of 29 = 1, 29
Common factors of 18 and 29 = 1
Can not reach on both treasures because jump size of 1 not allowed.

(d)
Factors of 20 =1, 2, 4, 5, 10, 20
Factors of 55 = 1, 5, 11, 55
Common factors of 15 and 39 = 1, 5
By jump size 3 reach on both treasures.

If Grumpy place the treasures on 4 and 15 'or' 18 and 29 then Jumpy can not reach both the treasures.

Question
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 

Solution:
(a) 
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factors of 18 and 35 = 1
This pair of numbers is co-prime. 

(b) 
Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1,37
Common factors of 15 and 37 = 1
This pair of numbers is co-prime. 

(c) 
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 = 1, 5, 83, 415
Common factors of 30 and 415 = 1, 5
This pair of numbers is not co-prime. 

(d) 
Factors of 17 = 1, 17
Factors of 69 = 1, 3, 23, 69
Common factors of 17 and 69 = 1
This pair of numbers is co-prime. 

(e)
Factors of 81 = 1, 3, 9, 27, 81
Factors of 18 = 1, 2, 3, 6, 9, 18
Common factors of 17 and 69 = 1,3
This pair of numbers is not co-prime.

Question
While playing the ‘idli-vada’ game with different number pairs, Anshu observed something interesting!
a. Sometimes the first common multiple was the same as the product of the two numbers.
b. At other times the first common multiple was less than the product of the two numbers.
Find examples for each of the above. How is it related to the number pair being co-prime?

Solution:
(a)
Example 
(i) First common factor of 9 and 4 =36
(ii) First common factor of 8 and 15 = 120
In (i) and (ii) pairs of number are co prime.
Hence first common factors of co prime numbers is equal to their product.

(b)
Example
(i) First common factor of 4 and 10 = 20
(ii) First common factor of 15 and 20 = 60
In (i) and (ii) pairs of number are not co prime.
Hence First common factor of non co prime numbers is less than their product.

(Note: You can take different pair of numbers)

Co-prime Art

Make such pictures for the following:
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:
(a)


(b)


(c)



(d)




Topic 5.4 Prime Factorisation

When we write any composite number in the product of prime numbers are called prime factorisation.
Example: 
20 = 2 × 2 × 5
24 = 2 × 2 × 2 × 3

Figure it Out

Question 1
Find the prime factorisation 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 number Which prime factorisation has one 2, two 3s, and one 11 = 2 × 3 × 3 × 11 = 198

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

Solution: 
 First we find the prime factorisation of 1955
Prime factorisation of 1955 = 5 × 17 × 23
Hence three number less than 30,whose product is 7955 are 5, 17, 23.

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

Solution:
(a)
⇒56 × 25 = 2 × 2 × 2 × 7 × 5 × 5

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

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

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 number are equal to product of first three prme number.
Hence required number = 2 × 3 × 5 = 30

(b)
The smallest number whose prime factorisation has four different prime number are equal to product of first four prime number.
Hence required number = 2 × 3 × 5 × 7 = 210

Using prime factorisation to check if two numbers are co-prime and Using prime factorisation to check if one number is divisible by another

Figure it Out

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)
Prime factorisation of
30 = 2 × 3 × 5
45 = 3 × 3 × 5
Here 3 and 5 are common prime factors of both the numbers therefore 30 and 45 are not co-prime.

(b)
Prime factorisation of
57 = 3 × 19
85 = 5 × 17
There  are no common prime factors of both the numbers therefore 57 and 85 are  co-prime.

(c)
Prime factorisation of
121 = 11 × 11
1331 = 11 × 11 × 11
Here 11 is common prime factors of both the numbers therefore 121 and 1331 are not co-prime.

(d)
Prime factorisation of
343 = 7 × 7 × 7
216 = 2 × 2 × 2 × 3 × 3 × 3
There  are no common prime factors of both the numbers therefore 343 and 216 are  co-prime.

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)
Prime factorisation of 
225 = 3 × 3 × 5 × 5
27 =  3 × 3 × 3
⇒225 is not divisible by 27 because the prime factorisation of 27 is not included in the prime factorisation of 225.

(b)
Prime factorisation of 
96 = 2 × 2 × 2 × 2 × 3
24 =  2 × 2 × 2 × 3
⇒96 is divisible by 24 because the prime factorisation of 24 is  included in the prime factorisation of 96.

(c)
Prime factorisation of 
343 = 7 × 7 × 7
17 =  1 × 17 
⇒343 is not divisible by 17 because the prime factorisation of 17 is not included in the prime factorisation of 343.

(d)
Prime factorisation of 
999 = 3 × 3 × 111
99  = 3 × 3 × 11
⇒999 is not divisible by 99 because the prime factorisation of 99 is not included in the prime factorisation of 999.

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:
Prime factorisation of first number = 2 × 3 × 7
Prime factorisation of second number = 3 × 7 × 11.
⇒ Here 3 is common prime  therefore both numbers are not co-prime.
⇒ No,one number is not divisible by other number because the prime factorisation of one number is not included in the prime factorisation of Second number.

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

Solution:
We know that co prime numbers have 1 as a common factors, when we find common factors of prime numbers we will get 1.Therefore he is right “Any two prime numbers are co-prime”.

Topic 5.5 Divisibility Tests


Question
Consider this statement:Numbers that are divisible by 10 are those that end with ‘0’. Do you agree?

Solution:
Multiples of 10 = 10, 20, 30,....., 100, 110,..., 200, 210,.....
Ecah multiple of 10 has 0 at its unit place therefore, Numbers that are divisible by 10 are those that end with‘0’.

(Note: Number divisible by 10 = Multiples of 10.)

Question
Explore by listing down the multiples: 5, 10, 15, 20, 25, ... What do you observe about these numbers? Do you see a pattern in the last digit?

Solution:
Each multiple of 5 has either a '0' or a '5' at the end.

Question
What is the largest number less than 399 that is divisible by 5? Is 8560 divisible by 5?

Solution:
The largest number less than 399 that divisible by 5 = 395.

Last digit of number 8560 is '0' therefore this number is divisible by 5.

Question
Consider this statement: Numbers that are divisible by 5 are those that end with either a ‘0’ or a ‘5’. Do you agree? 

Solution:
Number divisible by 5 = Multiples of 5 = 5, 10, 15,.....,50, 55, 60..........,125, 130, 135,....
Here the last digit of every multiple is either a '0' or a '5' . Therefore, Numbers that are divisible by 5 are those that end with either a ‘0’ or a ‘5’.

Question
Consider this statement: Numbers that are divisible by 2 are those that end with ‘0’, ‘2’, ‘4, ‘6’ or ‘8’. Do you agree?

Solution:
Number divisible by 2 = Multiples of 2 = 2, 4, 6, 8, ...., 20, 22, 24, ...., 56, 58, 60, ...., 100, 102,....
Here the last digit of every multiple is '0' or '2' or  '4' or  '6' or '8'.Therefore, Numbers that are divisible by 2 are those that end with ‘0’, ‘2’, ‘4, ‘6’ or ‘8’.

Question
What are all the multiples of 2 between 399 and 411?

Solution:
All the multiples of 2 between 399 and 411 = 400, 402, 404, 406, 408 and 410.

Question
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe?

Solution:
Numbers between 330 and 340 that are divisible by 4 = 332, 336

Numbers between 1730 and 1740 that are divisible by 4 = 1732, 1736

Numbers between 2030 and 2040 that are divisible by 4 = 2032 , 2036

Question
Is 8536 divisible by 4?

Solution:
Last two digits of the given number = 36
36 ÷ 4 = 9
Last two digit of number is divisible by 4 therefore 8536 id divisible by 4.

Question
Consider these statements:
a. Only the last two digits matter when deciding if a given number is divisible by 4.
b. If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
c. If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Do you agree? Why or why not?

(a)
Number divisible by 4 = Multiples of 4 = 4, 8, 12, ...., 56, 60, ....., 100 ,104, 108, 112, 116, 120, 124....
Here last two digit of each multiple of 4 is divisible by 4.Therefore we agree with the given statement.

(b)
Yes,if the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
Example: 5536, 4528, 8972 are divisible by 4 because the last two digits is divisible by 4.


(c)
Yes, if the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Example: 2548, 1316, 4124 are divisible by 4 and their last two digits also divisible by 4.

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

Solution:
Numbers between 120 and 140 that are divisible by 8 = 128, 136

Numbers between 1120 and 1140 that are divisible by 8 = 1128, 1136

Numbers between 3120 and 3140 that are divisible by 8 = 3128, 3136

Question
Change the last two digits of 8560 so that the resulting number is a multiple of 8.

Solution:
We can replace last two digits of the number by the following digits 

Question
Consider this statement:
a. Only the last three digits matter when deciding if a given number is divisible by 8.
b. If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
c. If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Do you agree? Why or why not?

Solution:
(a)
Number divisible by 4 = Multiples of 8 = 8, 16 ,24 ,..., 96, 104,..., 1496, 1504,.....3264,3272,......
Here last three digits of each multiple of 8 is divisible by 8.Therefore we agree with the given statement.

(b)
Yes, if the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
Example: 5728, 4928, 8976 are divisible by 8 because the last three digits is divisible by 8.

(c)
Yes, if the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Example: 2548, 7168 are divisible by 8 and their last three digits also divisible by 8.


Figure it Out

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 born till now, which years were leap years?
b. From the year 2024 till 2099, how many leap years are there?

Solution:
(a)
Born year= 2010
Leap years from 2010 to till now = 2012, 2016, 2020 and 2024

(b)
Leap years from 2024 to till 2099 = 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092 and 2096
Number of leap years from 2024 to 2099 = 19


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

Solution:
The largest 4 digits palindromes divisible by 4 = 8888
The smallest 4 digits palindromes divisible by 4 = 2112

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)
Sum of two even numbers gives a multiple of 4: Some times true
Example:
6 + 2 = 4 (multiple of 4)
10 + 4 = 14 (not multiple of 4)

(b)
Sum of two odd numbers gives a multiple of 4 : Some times true
Example;
7 + 5 = 12 (multiple of 4)
9 + 5 = 14 (not 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: 

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:
If number divisible by 8 then number  also divisible by 2 and 4.
If number divisible by 10 numb er also divisible by 5.

Guna checked 14560 by 8 and 10 then she declared that it was divisible by all of them.

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.

Soluton:
Prime factorisation of 10000 = 2 × 2 × 2 × 5 × 5 × 5 = 8 × 125

Hence, product of 8 and 125 is 10000.

Topic 5.6 Fun with numbers

A Prime Puzzle

The figure on the left shows the puzzle. The figure on the right shows the solution of the puzzle. Think what the rules can be to solve the puzzle.

Rules
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.

Solution:




(note: Replace 660 with 66)


 

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Mathematics with solution

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