Practicing RBSE Class 6 Maths Solutions and Class 6 Maths Chapter 5 Prime Time Solutions Question Answer helps develop logical thinking and accuracy.
Prime Time Class 6 Solutions
Ganita Prakash Class 6 Chapter 5 Solutions Prime Time
Figure it Out (Page 108)
Idli-Vada Game
Children sit in a circle and play a game of numbers.
One of the children starts by saying ‘1’. The second player says ‘2’, and so on. But when it is the turn of 3, 6, 9,… (multiples of 3), the player should say ‘ idli ’ instead of the number. When it is the turn of 5, 10,… (multiples of 5), the player should say ‘vada’ instead of the number. When a number is both a multiple of 3 and a multiple of 5, the player should say ‘idli-vada’! If a player makes any mistake, they are out.
The game continues in rounds till only one person remains.
Question 1.
At what number is ‘idli-vada’ said for the 10th time?
Solution:
To find the 10th time ‘idli-vada’ will be called, we have to write the multiples of 3 and 5 which are common multiples of both
15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …………………….
Hence, the required number is 150, when players have to say ‘idli-vada’.
Question 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 called for multiples of 3. Multiples of 3 between 1 to 90 is 3, 6, 9, 12, 15, 18, ……….. 90. There are 30 such numbers.
This is why children will say idli 30 times.
(b) ‘Vada’ is called for multiple of 5. Multiples of 5 between 1 to 90 is 5, 10, 15, 20, ……… 90. There are 18 such numbers.
(c) ‘Idli-Vada’ is called for common multiples of 3 and 5, i.e. 15. Multiple of 15 are 15, 30, 45, 60, 75, 90. There are 6 such numbers between 1 to 90.
Question 3.
What if the game was played till 900? How would your answers change?
Solution:
There are 300 multiples of 3 from 1 to 900. There are 180 multiples of 5 from 1 to 900. There are 60 multiples of 15 from 1 to 900. So,
(a) ‘Idli’ will be said 300 times which also includes the ‘idli’ said in ‘idli-vada’.
(b) ‘Vada’ will be said 180 times which also includes the ‘vada’ said in ‘idli-vada’.
(c) ‘Idli-Vada’ will be said 60 times.
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, this figure related to the ‘idli-vada’. The figure below is of a game played till 60.
Figure it Out (Page 110)
Question 1.
Find all multiples of 40 that lie between 310 and 410.
Solution:
The 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) The common factor of 7, 14, 21, 28 and 35 is 7 which is less than 40. One of these numbers is 35 whose digits sum is 3 + 5 = 8. Hence, I am 35.
(b) Common multiples of 3 and 5 are 15, 30, 45, 60, 75 and 90 (which are less than 100). One of these numbers is 45 whose one digit is 1 more than the other.
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:
6 is the only perfect number between 1 and 10, because factors of 6 are 1, 2, 3, 6. Sum of these factors is 1 + 2 + 3 + 6 = 12 which is twice 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 35 and 50 = 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, 175, 200, …….
Multiples of 50 : 50, 100, 150, 200, 250, 300, ……..
Hence, the required numbers which 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:
If the number after 50 is called ‘idli-vada’ then it means the least common multiple (LCM) of both the numbers must be slightly greater than 50. The LCM of 6 and 9 is 54 which is the first common number after 50. Hence, the required numbers are 6 and 9.
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, 70
Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70
Common factors of 28 and 70 = 1, 2, 7 and 14
Hence, the required jump sizes are 1, 2, 7 and 14 to land on both the numbers.
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:
Here, instead of 6, multiples of 3 can also be taken. 24, 48 and 72 are common multiples of 3 and 8.
Question 9.
Find the smallest number that is a multiple of all the numbers from 1 to 10, except for 7.
Solution:
To find the common-multiple of all numbers from 1 to 10 except 7, we have to find the least common multiple of numbers from 1 to 10 (except 7).
Here, 1 = 1, 2 = 2, 3 = 3, 4 = 2 × 2, 5 = 5, 6 = 2 × 3 8 = 2 × 2 × 2, 9 = 3 × 3, 10 = 2 × 5
∴ Least common multiples (LCM) = 2 × 2 × 2 × 3 × 3 × 5 = 360
Hence, the required number is 360 which is a multiple of all the numbers from 1 to 10, except 7.
Question 10.
Find the smallest number that is multiple of all the numbers from 1 to 10.
Solution:
To find the common multiple of all the numbers from 1 to 10, we have to find the least common multiple of numbers from 1 to 10.
Here, 1 = 1, 2 = 2, 3 = 3, 4 = 2 × 2, 5 = 5, 6 = 2 × 3, 7 = 7, 8 = 2 × 2 × 2, 9 = 3 × 3, 10 = 2 × 5
∴ LCM = The product of the maximum number of prime factors other than other factors.
= 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520
Hence, the required smallest number that is a multiple of all the numbers from 1 to 10 is 2520.
Figure it Out (Page 114)
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 even prime number that satisfies the criteria of being a prime number. 2 can be divided only by 1 and by itself. All other even numbers are divisible by 1, 2 and themselves.
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:
List of prime numbers till 100 is : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71,73, 79, 83, 89, 97
Difference between two successive primes—
Minimum difference = 1 (between 2 and 3)
Maximum difference = 8 (between 88 and 97)
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, the number of prime numbers in each row was not the same, the number of prime numbers varies between rows. The range 90 to 99 has the least number of prime numbers with only one prime number being 97. The decades 0 to 9 and 10 to 99 have the largest number of prime numbers, with four prime numbers each.
Question 4.
Which of the following numbers are prime : 23, 51, 37, 26?
Solution:
23 and 37 have no divisors other than one and themselves. Hence, 23 and 37 are prime numbers here.
Question 5.
Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Solution:
Three pairs of prime numbers less than 20 whose sum is a multiple of 5 are (2, 3), (2, 13) and (7, 13).
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 up to 100 having the same digits : (13, 31), (17, 71), (37, 73) and (79, 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, 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 prime numbers other than 3 and 5, 17 and 19 between 1 and 100 are : (5, 7), (11, 13), (29, 31), (41, 43), (59, 61), (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 statement.
The units digit of a prime number is 1, 3, 7 or 9 (except 2). The numbers whose units digit is 0, 2, 4, 6, 8 are divisible by 2. Thus, there is no prime number whose units digit is 4.
(b) False statement.
When two or more prime numbers are multiplied, the result is always a composite number and not a prime number. For example, 2 × 3 = 6, 2 × 5 = 10 etc.
(c) False statement.
Prime numbers have exactly two factors : 1 and the number itself.
(d) False statement.
Number 2 is a even prime number. 2 is the only even number which is prime.
(e) True statement.
For every prime number greater than 2, the next number is composite. For example, 5 is a prime number and the next number 6 is a composite number. The next number always be even number which in always divisible by 2.
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 (Two different composite numbers)
60 = 2 × 2 × 3 × 5 (Three different composite numbers)
91 = 7 × 13 (Two different composite numbers)
105 = 3 × 5 × 7 (Three different composite numbers)
330 = 2 × 3 × 5 × 11 (Four different composite numbers)
Question 11.
How many three-digit prime numbers can you make using each of 2, 4 and 5 once?
Solution:
The digits 2, 4 and 5 cannot together form a prime number because when the unit digit is 2 or 4, it is divisible by 2 and when the unit digit is 5, the number is divisible by 5, So the digits 2, 4 and 5 cannot together form a prime number.
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:
Five prime numbers which when multiplied by 2 and then added 1 give another prime number.
(1) 2 (∵ 2 × 2 + 1 = 5)
(2) 3 (∵ 3 × 2 + 1 = 7)
(3) 5 (∵ 5 × 2 + 1 = 11)
(4) 11 (∵ 11 × 2 + 1 = 23)
(5) 23 (∵ 23 × 2 + 1 = 47)
Figure it Out (Page 120)
Question 1.
Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.
Solution:
| Number | Prime factorisation |
| 64 | 2 × 2 × 2 × 2 × 2 × 2 |
| 104 | 2 × 2 × 2 × 13 |
| 105 | 3 × 5 × 7 |
| 243 | 3 × 3 × 3 × 3 × 3 |
| 320 | 2 × 2 × 2 × 2 × 2 × 2 × 5 |
| 141 | 3 × 47 |
| 1728 | 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 |
| 729 | 3 × 3 × 3 × 3 × 3 × 3 |
| 1024 | 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 |
| 1331 | 11 × 11 × 11 |
| 1000 | 2 × 2 × 2 × 5 × 5 × 5 |
Question 2.
The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?
Solution:
To get the required number, we have to multiply the given prime numbers.
2 × 3 × 3 × 11 = 198
Hence, the required numbers = 198
Question 3.
Find three prime numbers, all less than 30, whose product is 1955.
Solution:
The prime factorisation of 1955 is 5 × 17 × 23. All the factors are prime numbers and are less than 30. Hence, the required prime numbers whose product is 1955 are 5, 17 and 23.
Question 4.
Find the prime factorisation of these numbers without multiplying first
(a) 56 × 25
(b) 108 × 75
(c) 1000 × 81
Solution:
(a) Prime factors of 56 = 2 × 2 × 2 × 7
Prime factors of 25 = 5 × 5
So the prime factorisation of 56 × 25 = 2 × 2 × 2 × 5 × 5 × 7
(b) Prime factors of 108 = 2 × 2 × 3 × 3 × 3
Prime factors of 75 = 3 × 5 × 5
So the prime factorisation of 108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
(c) Prime factors of 1000 = 2 × 2 × 2 × 5 × 5 × 5
Prime factors of 81= 3 × 3 × 3 × 3
Prime factorisation of 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 is 2 × 3 × 5 = 30
(b) The smallest number whose prime factorisation has four different prime numbers is 2 × 3 × 5 × 7 = 210
Figure it Out (Page 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) Prime factors of 30 = 2 × 3 × 5
Prime factors of 45 = 3 × 3 × 5
Common factors of both numbers = 3 × 5 = 15
Hence, 30 and 45 are not a pair of co-prime numbers.
(b) Prime factors of 57 = 3 × 19
Prime factors of 85 = 5 × 17
There is no common factors of 57 and 85 other than 1, hence, 57 and 85 are a pair of co-prime numbers.
(c) Prime factors of 121 = 11 × 11
Prime factors of 1331 = 11 × 11 × 11
Common factors of 121 and 1331 = 11 × 11 = 121
Hence, 121 and 1331 are not a pair of co-prime numbers.
(d) Prime factors of 343 = 7 × 7 × 7
Prime factors of 216 = 2 × 2 × 2 × 3 × 3 × 3
There is no common factors other than 1, hence 343 and 216 are a pair of 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) Prime factorisation of225 = 3 × 3 × 5 × 5
Prime factorisation of 27 = 3 × 3 × 3
Since 225 contains 3 × 3 and does not have enough factors of 3 to match 3 × 3 × 3.
So 225 does not have sufficient factors to be divisible by 27.
Hence, 225 is not divisible by 27.
(b) Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3
Prime factorisation of 24 = 2 × 2 × 2 × 3
Since 96 includes the some factors of 24. i.e. 2 × 2 × 2 × 3
Hence, 96 is divisible by 24.
(c) Prime factorisation of 343 = 7 × 7 × 7
Prime factorisation of 17 = 1 × 17
Since the prime factorisation of 343 includes the prime factor 7 but not 17, so the 343 is not divisible by 17.
(d) Prime factorisation of 999 = 3 × 3 × 3 × 37
Prime factorisation of 99 = 3 × 3 × 11
Since 999 does not contain the factor 11, so 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 and the second number 3 × 7 × 11 have common factors 3 × 7. Both numbers are divisible by 3 × 7 . So they are not co-prime. Since neither number contains all the factors of the other nor can divide the other.
Question 4.
Guna says, “Any two prime numbers are co-prime?”. Is he right?
Solution:
Yes. Guna is right. Any two prime numbers are co-prime as they do not have common factor other than 1. Therefore, any two prime numbers are always co-prime. For example: 5 and 7, 7 and 11.
Figure it Out (Page 125)
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) Do yourself. For example, if your birth year 2010. then there are 4 leap years from 2010 to 2024 : 2012, 2016, 2020 and 2024.
(b) There are 19 leap years from 2024 to 2099. These are : 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096.
Question 2.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Solution:
Largest 4-digit number divisible by 4 and is also palindrome = 8888
Smallest 4-digit number divisible by 4 and is also palindrome = 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′. This statement is sometimes true.
For example : 2 + 6 = 8 is a multiple of 4, while 6 + 4 = 10 is not a multiple of 4.
(b) ‘Sum of two odd numbers gives a multiple of 4’. This statement is sometimes true. For
example : 1 + 7 = 8 is a multiple of 4, while 3 + 7 = 10 is not a multiple of 4.
Question 4.
Find the remainders obtained when each of the following r Jmbers are divided by (a) 10, (b) 5, (c) 2.
78, 99, 173, 572, 980, 1111, 2345
Solution:
(i) 78 : Here we are to divide 78 by 10. 5 and 2.
The students should do the rest themselves.
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 a number is divisible by 8 then it is automatically be divisible by 2 and 4. Similarly, if a number is divisible by 10, then that number will be divisible by 2 and 5. So, Guna checked for divisibility of 14560 by only 8 and 10 then declared that it was also 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:
Only 5600, 6000 and 77622160 are divisible of 2, 4, 5, 8 and 10.
Question 7.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Solution:
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
So, 2 × 2 × 2 × 2 = 16
5 × 5 × 5 × 5 = 625
So, 16 and 625 are the required number whose product is 10000.
Prime Time Class 6 Question Answer
Prime Time Class 6 Extra Questions
Multiple Choice Questions—
Question 1.
Which of the following is a composite number?
(a) 5
(b) 13
(c) 12
(d) 23
Answer:
(c) 12
Question 2.
The numbers whose factors are 1 and the number itself are called—
(a) Prime numbers
(b) Composite numbers
(c) Co-prime numbers
(d) Semi-composite numbers
Answer:
(a) Prime numbers
Question 3.
Which of the following is a number divisible by 3—
(a) 29
(b) 71
(c) 81
(d) 91
Answer:
(c) 81
Question 4.
All the factors of 6 are—
(a) 2, 3, 6
(b) 1, 2, 3, 6
(c) 0, 1, 2, 3, 6
(d) 2, 4, 6
Answer:
(b) 1, 2, 3, 6
Question 5.
4 and 18 have common factors—
(a) 2, 4, 6, 9
(b) 2, 4, 18
(c) 2
(d) 1, 2
Answer:
(d) 1, 2
Question 6.
Which of the following is a prime number?
(a) 23
(b) 18
(c) 25
(d) 15
Answer:
(a) 23
Question 7.
Which of the following is a prime factorisation of 54?
(a) 2 × 27
(b) 2 × 3 × 9
(c) 54 × 1
(d) 2 × 3 × 3 × 3
Answer:
(d) 2 × 3 × 3 × 3
Question 8.
LCM of 12 and 18 will be—
(a) 12
(b) 36
(c) 54
(d) 72
Answer:
(b) 36
Question 9.
Which of the following is not a prime number—
(a) 2
(b) 1
(c) 3
(d) 5
Answer:
(b) 1
Question 10.
The number which is divisible by 2 will be—
(a) 120
(b) 115
(c) 313
(d) 523
Answer:
(a) 120
Fill in the blanks—
1. A number is a ………………….. of each of its factors.
Answer:
multiple
2. Those numbers which have more than two factors are called ………………….. numbers.
Answer:
composite
3. Natural numbers which are exactly divisible by two are called ………………….. numbers.
Answer:
even
4. The LCM of 2 or more numbers will be the smallest common ………………….. .
Answer:
multiples
5. The numbers 6 and 28 are ………………….. numbers.
Answer:
perfect
6. The number of multiples of a given number is ………………….. .
Answer:
infinite
7. Every number is a multiple of ………………….. .
Answer:
itself
8. Every multiple of a given number is ………………….. that number.
Answer:
greater than or equal to
9. ………………….. is a factor of every number.
Answer:
1
10. Every factor of a number is ………………….. of that number.
Answer:
divisor.
Write True/False for the following statements—
1. The sum of three odd numbers is even. (True/False)
Answer:
False
2. The sum of two odd numbers and an even number is even. (True/False)
Answer:
True
3. The product of three odd numbers is odd. (True/False)
Answer:
True
4. If an even number is divided by 2, then the quotient is always odd. (True/False)
Answer:
False
5. All prime numbers are odd. (True/False)
Answer:
False
6. Prime numbers have no factors.
(True/False)
Answer:
False
7. All even numbers are composite numbers.
(True/False)
Answer:
False
Make the right match—
Question 1.
| 1. 35 | (a) Multiple of 8 |
| 2. 15 | (b) Divisor of 20 |
| 3. 16 | (c) Divisor of 50 |
| 4. 20 | (d) Multiple of 7 |
| 5. 25 | (e) Divisor of 30 |
Answer:
1. – (d), 2. – (e), 3. – (a). 4. – (b), 5. – (c).
| 1. 35 | (d) Multiple of 7 |
| 2. 15 | (e) Divisor of 30 |
| 3. 16 | (a) Multiple of 8 |
| 4. 20 | (b) Divisor of 20 |
| 5. 25 | (c) Divisor of 50 |
Very Short Answer Type Questions—
Question 1.
Write all the factors of 36.
Solution:
All factors of 36—1, 2, 3, 4, 6, 9, 12, 18 and 36.
Question 2.
Write the first five multiples of 7.
Solution:
7 × 1 = 7,
7 × 2 = 14,
7 × 3 = 21,
7 × 4 = 28,
7 × 5 = 35
Therefore, the first five multiples of 7 are 7, 14, 21, 28 and 35.
Question 3.
What is called a perfect number?
Solution:
A number whose sum of all its factors is double of that number is called a perfect number.
Question 4.
Write the first five multiples of 6.
Solution:
6 × 1 = 6,
6 × 2 = 12,
6 × 3 = 18,
6 × 4 = 24,
6 × 5 = 30
Therefore, the first five multiples of 6 are 6, 12, 18, 24 and 30.
Question 5.
Define prime numbers.
Solution:
The numbers whose factors are 1 and the number itself are called prime numbers.
Question 6.
What are composite numbers?
Solution:
Those numbers which have more than two factors are called composite numbers.
Question 7.
If a number is divisible by both 2 and 3, then by which number is that number also divisible?
Solution:
That number is also divisible by 6.
Question 8.
Write the prime number 53 as the sum of three prime numbers.
Solution:
53 = 3 + 19 + 31
Question 9.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution:
3 and 4 are co-prime numbers, so the common multiples of 3 and 4 are the multiples of 3 × 4 = 12.
∴ The required numbers which are less than 100 and common multiples of 3 and 4 are:
12, 24, 36, 48, 60, 72, 84 and 96.
Question 10.
A number is divisible by 5 and 12 both. By which other number will it always be divisible?
Solution:
Since the number is divisible by 5 and 12 both, and 5 and 12 are co-prime numbers.
Therefore, this number will also be divisible by 5 × 12 = 60.
Short Answer Type Questions—
Question 1.
Find the common factors of 8, 12 and 20.
Solution:
Factors of 8 : 1, 2, 4 and 8
Factors of 12 : 1, 2, 3, 4, 6 and 12
Factors of 20 : 1, 2, 4, 5, 10 and 20
Obviously, the common factors of 8, 12 and 20 are 1, 2 and 4.
Question 2.
Find the LCM of 20, 25 and 30.
∴ LCM of 20, 25 and 30 = 2 × 2 × 3 × 5 × 5 = 300
Question 3.
Write four multiples of the given numbers.
(i) 7, ……………, ……………, ……………, ……………,
(ii) 13, ……………, ……………, ……………, ……………,
(iii) 17, ……………, ……………, ……………, ……………,
(iv) 19, ……………, ……………, ……………, ……………,
Solution:
(i) 7, 14, 21, 28, 35
(ii) 13, 26, 39, 52, 65
(iii) 17, 34, 51, 68, 85
(iv) 19, 38, 57, 76, 95
Question 4.
Check divisibility of 67,527 by. 9 without division.
Solution:
Sum of digits of 67, 527 = 6 + 7 + 5 + 2 + 7 = 27
which is exactly divisible by 9.
Therefore, the number 67,527 is exactly divisible by 9.
Question 5.
Find all the prime factors of 1729 and arrange them rn ascending order. Now write the relation, if any, between two consecutive prime factors.
Solution:
∴ Prime factorisation of 1729 = 7 × 13 × 19.
Obviously, the difference between two consecutive prime factors is 6.
Question 6.
The product of three consecutive numbers is always divisible by 6. Explain this statement with the help of some examples.
Solution:
Product of some three consecutive numbers:
1 × 2 × 3 = 6 11 × 12 × 13 = 1716
2 × 3 × 4 = 24 17 × 18 × 19 = 5814
3 × 4 × 5 = 60 20 × 21 × 22 = 9240
In each product unit place digit is 6, 4 or 0. So each product is divisible by 2 and the sum of the digits of these products are 6, 9, 12, 15 and 18 which is divisible by 3. So each product is divisible by 3.
As 2 and 3 are co-prime. 2 × 3 = 6 will divide each of the above products. Therefore, the product of three consecutive numbers is always divisible by 6.
Question 7.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement by taking some examples.
Solution:
The sum of two consecutive odd numbers :
3 + 5 = 8, 5 + 7 = 12,
11 + 13 = 24, 23 + 25 = 48,
51 + 53 = 104, 69 + 71 = 140
Clearly, every sum of two consecutive odd numbers given abpve is divisible by 4. So it is true that the sum of two consecutive odd numbers is divisible by 4.
Essay Type Questions—
Question 1.
Find the smallest number which when divided by 9, 15, 25 leaves remainder 4 in every case.
Solution:
Since here we have to find the smallest number which after dividing by 9, 15, 25 leaves remainder 4. So by LCM we will first find the smallest number which can be fully divided by these numbers.
LCM = 5 × 3 × 3 × 5 = 225
By adding 4 more to this number, we will get the required number.
So 225 + 4 = 229
Question 2.
Four clocks ring after 5, 8, 10, 20 minutes respectively. After how much time of ringing together, will all the four clocks ring together again?
Solution:
LCM of 5, 8, 10, 12 is
∴ LCM = 2 × 2 × 2 × 3 × 5 = 120
All four clocks will ring together again after 120 minutes i.e. 2 hours.
Question 3.
Renu buys two fertilizer bags weighing 75 kg and 69 kg. Find the maximum value of the weight that can measure the weight of both the bags exactly.
Solution:
75 kg and 69 kg fertilizer bags are to be measured. Therefore, the weight should be such that it completely divides the bags of both capacities. Also, its value should
be maximum. Therefore, the maximum weight will be the HCF of 75 and 69.
∴ 69 = 3 × 23 and 75 = 3 × 5 × 5
Hence, the HCF of 69 and 75 = 3
∴Maximum value of weight = 3 kg
Question 4.
Three boys start walking from the same place by taking steps together. The measure of their steps is 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each of them should cover so that the distance is covered in complete steps?
Solution:
Three boys start from the same place together, and the measure of their steps is 63 cm, 70 cm and 77 cm respectively. To find the minimum and equal distance covered by each of them in complete steps, we will take the LCM of 63, 70 and 77.
∴ Required LCM = 2 × 3 × 3 × 5 × 7 11 = 6930 cm
∴ Minimum distance covered by each so that the distance is covered in full steps
= 6930 cm
= 69.30 m
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