By using Ganita Prakash Book Class 6 Solutions and Chapter 5 Prime Time Class 6 NCERT Solutions Question Answer, students can improve their problem-solving skills.
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 2)
Idli-Vada Game
Children sit in a circle and play a game of numbers. One of the children starts by saying T. The second player says ‘2’, and so on. But when it si the turn of 3, 6, 9,… (multiple 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.
For which numbers should the players say ‘idli’ instead of saying the number? These would be 3, 6, 9, 12, 18,… and so on.
These would be 5, 10, 20,… and so on.
Which is the first number for which the players should say, ’idli-vada’? It is 15, which is out of other such numbers that are multiple of 5. Find out other such numbers that are multiples of both 3 and 5. These numbers are called common multiples.
Question 1.
At what number is ‘idli-vada’ said for the 10th time?
Solution:
It is given that ‘idli-vada is said for the number which is both a multiple of 3 and a multiple of 5 i.e. multiple of 15.
So, the 10th multiple of 15 is 150.
Thus, ‘idli-vada’ is said for 10th time for the number 150.
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:
There are 300 multiples of 3 between 1 and 900 and there are 180 multiples of 5 between 1 and 900. There are 60 multiples of 15 between 1 and 900.
(a) “idli” is said: 300 times (including the times “idli-vada” is said).
(b) “vada” is said: 180 times (including the times “idli-vada” is said).
(c) “idli-vada” is 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 is related to the ‘idli-vada’ game
The figure if the game is played till 60 is given below.
Figure it Out (Page 110)
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:
The only perfect number between 1 and 10 is 6.
- Proper divisors are 1, 2, 3, 6
- Sum of proper divisors: 1 + 2 + 3 + 6 = 12
- 12 is twice of 6, hence 6 is a perfect number.
Question 4.
Find the common factors of:
(a) 20 and 28
(b) 35 and 50
(0 4, 8 and 12
(d) 5, 15 and 25
Solution:
(a) Factors of 20 are 1, 2, 4, 5, 10, 20 Factors of 28 are 1, 2, 4, 7, 14, 28 Common factors are 1, 2, 4
(b) Factors of 35 are 1, 5, 7, 35 Factors of 50 are 1, 2,’5, 10, 25, 50 Common factors are 1,5
(c) Factors of 4 are 1, 2, 4 Factors of 8 are 1, 2, 4, 8 Factors of 12 are 1, 2, 3, 4, 6, 12 Common factors are 1, 2, 4
(d) Factors of 5 are 1, 5 Factors of 15 are 1, 3, 5, 15 Factors of 25 are 1, 5, 25 Common factors are 1,5.
Question 5.
Find any three numbers that are multiples of 25 but not multiples of 50.
Solution:
The multiples of 25 are 25, 50, 75, 100, 125, 150,………
The multiple of 50 are 50, 100, 150, 200, 250, 300,………..
Three numbers that are multiples of 25 but not multiples of 50 are 25, 75, and 125.
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:
‘Idli’ said for multiple of 3 at 3, 6, 9, 12,……….
‘Vada’ said for multiple of 5 at 5, 10, 15, 20,……….
The first time anybody says ‘idli-vada’ after the number 50 is 60.
Question 7.
In the treasure-hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both numbers?
Solution:
The factor of 28 are 1, 2, 4, 7, 14, 28.
The factor of 70 are 1, 2, 5, 7, 10, 14, 35, 70.
So, the jump sizes landing on both numbers are 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:
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 smallest number that is a multiple of all
numbers from 1 to 10 except for 7, we need to determine LCM of the numbers from 1 to 10 except 7.
Prime factorisation of numbers from 1 to 10.. ”
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
LCM (1 to 10 excepts 7) = 23 × 32 × 5 = 360
Thus, the required smallest number is 360. y
Question 10.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Solution:
To find the smallest number that is divisible by all the numbers from 1 to 10 (both inclusive), let us first find the LCM of all the numbers between 1 and 10 (both inclusive)
we have
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 = Product of the highest power of the prime factors including other factors = 2 × 2 × 2 × 3 × 3 × 5 × 7
= 8 ×9 × 5 × 7 = 2520
The smallest number that is a multiple of all the numbers from 1 to 10 is 2520
5.2 Prime Numbers 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 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 (See NCERT Textbook, Page 113)? Which decades have the least number of primes? Which have the most number of primes?
Solution:
There is not an equal number of primes in every row. The number of primes varies between rows. The decade 90-99 has the least number of primes with only 1 prime (97).
The decades 0-9 and 10-19 have the greatest number of primes, each with 4 primes.
Question 4.
Which of the following numbers are prime?
23, 51, 37, 26
Solution:
23 and 37 have no divisors other than 1 and themselves, making them prime.
Question 5.
Write three pairs of prime numbers less than 20 • whose sum is a multiple of 5.
Solution:
The pairs are less than 20 whose sum is a multiple of 5 are (2, 3), (3, 7), (7, 13), (3, 17), (13, 17), and (2, 13).
Question 6.
The numbers 13 and 31 are prime numbers. Both these numbers have the same digits 1 and 3. Find such pairs of prime numbers up to 100.
Solution:
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:
The factors of 105 are 3 × 5 × 7.
Question 11.
How many three-digit prime numbers can you make using each of 2, 4 and 5 once?
Solution:
The three-digit numbers using 2,4 and 5 once are
245 (divisible by 5) → Not prime
254 (divisible by 2) → Not prime
425 (divisible by 5) → Not prime
452 (divisible by 2) → Not prime
524 (divisible by 2) → Not prime
542 (divisible by 2) → No prime
Thus, we cannot make three-digit prime numbers 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:
There are other primes for which doubling and adding 1 gives another prime:
2 × 5 + 1 = 11
2 × 11 + 1 = 23
2 × 23 + 1 = 47
2 × 29 + 1 = 59
2 × 41 + 1 = 83
2 × 53 + 1 = 107
5.3 Co-prime Numbers for Safekeeping Treasures 5.4 Prime Factorisation 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:
Prime factorisation of the given numbers are
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 and
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:
Prime factorisation of a number = 2 × 3 × 3 × 11
∴ Number = 2 × 3 × 3 × 11 = 198
Question 3.
Find three prime numbers, all less than 30, whose product is 1955.
Solution:
Three prime numbers, all less than 30, whose product is 1955 are 5, 17 and 23.
i. e. 1955 = 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:
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
Figure it Out (Page 122)
Question 1.
Are the following pairs of numbers co-prime? Guess first and then use prime factorization to verify your answer.
(a) 30 and 45
(b) 57 and 85
(c) 121 and 1331
(d) 343 and 216
Solution:
(a) The factors of 30 are 2 × 3 × 5 and 45 are 3 × 3 × 5.
So, there are two common prime factors 3 × 5.
Hence, it is not a co-prime number.
(b) The factors of 57 are 3 × 19 and 85 are 5 × 17.
So, there is no common prime factor.
Hence, it is a co-prime number.
(c) The factors of 121 are 11 × 11 and 1331 are 11 × 11 × 11.
So, there is a common factor 11.
Hence, it is not a co-prime number.
(d) The factors of 343 are 7 × 7 × 7 and 216 are 6 × 6 × 6.
So, there is no common prime factor.
Hence, it is a co-prime number.
Question 2.
Is the first number divisible by the second? Use prime factorization.
(a) 225 and 27
(b) 96 and 24
(c) 343 and 17
(d) 999 and 99
Solution:
(a) Prime Factors of 225 and 27:
225 = 3 × 3 × 5 × 5 and 27 = 3 × 3 × 3
Since 225 contains 3×3 and does not have enough factors of 3 to match 3×3 × 3, 225 does not have sufficient factors to be divisible by 27.
Therefore, 225 is not divisible by 27.
(b) Prime Factors of 96 and 24:
96 = 2 × 2 × 2 × 2 × 2 × 3 and 24 = 2 × 2 × 2 × 3.
Since 96 includes the required factors to match those in 24, it is divisible by 24.
(c) Prime Factors of 343 and 17:
343 = 7 × 7 × 7 and 17 = 1 × 17
Since the prime factorization of 343 contains the prime factor 7 and not 17, 343 is not divisible by 17.
Thus, 343 is not divisible by 17.
(d) Prime Factors of 999 and 99: t
999 = 3 × 3 × 3 × 37 and 99 = 3 × 3 × 11
Since 999 does not include the factor 11 required for 99, 999 is not divisible by 99.
Question 3.
The first number has prime factorization 2 × 3 × 7 and the second number has prime factorization 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
No, because 3 and 7 are common prime factors in both numbers according to the definition of co-prime numbers. Also, 2 is a prime factor of the first number but not a prime factor of the second number, and 11 is the prime factor of the second but not a prime factor of the first. Therefore, one of them does not divide the other number.
Question 4.
Guna says, “Any two prime numbers are co-prime”. Is he right?
Solution:
Yes, he is right, Because any two prime numbers do not have any common factors. Therefore, they are co-prime.
5.5 Divisibility Tests Divisibility by 10 (Page 123)
Question 1.
Numbers that are divisible by 10 are those that end with ‘0’. Do you agree?
Solution:
Yes, I agree. A number is divisible by 10 if its last digit is ‘0’. This is because 10 is made up of the factors 2 and 5, and for a number to be divisible by 10, it must have both 2 and 5 as factors. If the number ends with ‘0’, it means that both conditions are met, making the number divisible by 10.
Divisibility by 5 (Page 123)
Question 2.
Numbers that are divisible by 5 are those that end with either a ‘0’ or a ‘5’. Do you agree?
Solution:
Yes, I agree. A number is divisible by 5 if its last digit is either ‘0’ or ‘5’. This is because 5 is a factor of both 10 and 5, and for a number to be divisible by 5, it must have 5 as a factor. Numbers ending in ‘0’ or ‘5’ satisfy this condition, making them divisible by 5.
Divisibility by 2 (Page 124)
Question 1.
Numbers that are divisible by 2 are those that end with ‘0’, ‘2’, ‘4, ‘6’ or ‘8’. Do you agree? What are all the multiples of 2 between 399 and 411?
Solution:
Yes, I agree. A number is divisible by 2 if its last digit is ‘0’, ‘2’, ‘4’, ‘6’, or ‘8’, because such numbers are even and can be divided evenly by 2.
The multiples of 2 between 399 and 411 are: 400, 402, 404, 406, 408, and 410.
Divisibility by 4 (Page 124)
Question 2.
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:
To find numbers divisible by 4, we check if the number is divisible by 4 by looking at the last two digits. If the number formed by the last two digits is divisible by 4, the whole number is divisible by 4.
- Between 330 and 340: The numbers divisible by 4 are: 332, 336, and 340.
- Between 1730 and 1740: The numbers divisible by 4 are: 1732, 1736, and 1740.
- Between 2030 and 2040: The numbers divisible by 4 are: 2032, 2036, and 2040.
Observation: In all these ranges, the numbers divisible by 4 follow a consistent pattern, increasing by 4 between consecutive multiples.
Question 3.
Is 8536 divisible by 4?
Solution:
Yes, 8536 is divisible by 4. To check, we look at the last two digits, which are 36. Since 36 is divisible by 4. the entire number 8536 is divisible by 4.
Question 4.
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?
Answer:
(a) Agree. The divisibility rule for 4 states that if the number formed by the last two digits is divisible by 4, then the entire number is divisible by 4. Therefore, we only need to focus on the last two digits.
(b) Agree. If the last two digits form a number divisible by 4, it guarantees that the entire number is divisible by 4, as the divisibility rule for 4 depends on these last two digits.
(c) Agree. If a number is divisible by 4, then its last two digits must also form a number divisible by 4. This is a consequence of the divisibility rule for 4.
Divisibility by 8 (Page 125)
Question 1.
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?
Answer:
To find numbers divisible by 8, we check if the number formed by the last three digits is divisible by 8.
- Between 120 and 140: The numbers divisible by 8 are 120, 128, and 136.
- Between 1120 and 1140: The numbers divisible by 8 are 1120, 1128 and 1136.
- Between 3120 and 3140: The numbers divisible by 8 are 3120, 3128 and 3136.
Observation: In each range, the numbers divisible by 8 appear consistently, following a pattern where the difference between consecutive multiples of 8 is 8. Moreover, the last three digits of numbers in different ranges behave similarly in determining divisibility by 8.
Question 2.
Change the last two digits of 8560 so that the resulting number is a multiple of 8.
Answer:
To make 8560 a multiple of 8, we need to change the last two digits (60) so that the resulting number is divisible by 8.
The closest multiples of 8 near 8560 are 8568 and 8552. Therefore, the possible numbers are: 8552, 8568
Both 8552 and 8568 are divisible by 8.
Question 3.
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?
Answer:
(a) Agree. The divisibility rule for 8 states that if the number formed by the last three digits is divisible by 8, then the entire number is divisible by 8. Therefore, only the last three digits matter.
(b) Agree. If the last three digits form a number divisible by 8, this guarantees that the entire number is divisible by 8. This follows directly from the divisibility rule for 8.
(c) Agree. If the original number is divisible by 8, then the number formed by the last three digits must also be divisible by 8. This is because the entire number satisfies the condition for divisibility by 8, and the last three digits must reflect that.
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 bom till now, which years were leap years?
(b) From the year 2024 till 2099, how many leap years are there?
Solution:
(a) Let the year you were born be 2010.
2010 (not divisible by 4)
2011 (not divisible by 4)
2012 (divisible by 4)
2013 (not divisible by 4)
2014 (not divisible by 4)
2015 (not divisible by 4)
2016 (divisible by 4)
2017 (not divisible by 4)
2018 (not divisible by 4)
2019 (not divisible by 4)
2020 (divisible by 4)
2021 (not divisible by 4)
2022 (not divisible by 4)
2023 (not divisible by 4)
2024 (divisible by 4)
Thus, the years that were leap years are 2012, 2016, 2020 and 2024.
(b) Since, leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. Therefore, the leap years from 2024 till 2099 are 2024,2028,2032, 2036, 2040,2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076,2080, 2084, 2088,2092 and 2096.
Hence, there are 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.
(i) The sum of two even numbers gives a multiple of 4.
(ii) The sum of two odd numbers gives a multiple of 4.
Solution:
(i) The sum of two even numbers gives a multiple of 4.
This statement is sometimes true.
Example: when it is true; 10 + 18 = 28 is a multiple of 4.
When it is false; 10 + 16 = 26 is not a multiple of 4.
(ii) The sum of two odd numbers gives a multiple of 4.
This statement is sometimes true.
Example: when it is true; 3 + 5 = 8 is a multiple of 4.
When it is false; 7 + 11 = 18 is not a multiple of 4.
These statements are always true:
The sum of two even numbers is always even.
Example: 2 + 4 = 6, 16 + 28 = 44.
The sum of two odd numbers is always even.
Example: 3 + 5 = 8, 7 + 19 = 26
The multiple of odd numbers and even numbers is always even.
Example: 2 × 3 = 6, 7 × 8 = 56
The statement is sometimes true:
Multiple rational and irrational numbers are sometimes rational and sometimes irrational numbers.
Example: √3 × 2 = 2√3, √3 × 0 = 0
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) We know that a number is divisible by 10 if and only if its last digit is 0.
For 78, the last digit is 8.
Therefore, when 78 is divided by 10, the remainder is 8. For 99, the last digit is 9. Therefore, when 99 is divided by 10, the remainder is 9.
For 173, the last digit is 3. Therefore, when 173 is divided by 10, the remainder is 3.
For 572, the last digit is 2. Therefore, when 572 is divided by 10, the remainder is 2.
For 980, the last digit is 0. Therefore, remainder is 0. For 1111, the last digit is 1. Therefore, the remainder is 1.
For 2345, the last digit is 5. Therefore, the remainder is 5.
(ii) We know that a number is divisible by 5 if it either ends with a ‘0’ or a ‘5’.
For 78, the last digit is 8. When 8 is divided by 5, the remainder is 3.
For 99, the last digits 9. When 9 is divided by 5, the remainder is 4.
For 173, the last digit is 3. When 3 is divided by 5, the remainder is 3. [∵ 3 < 5]
For 572, the last digit is 2. When 2 is divided by 5 the remainder is 2. [∵ 2 < 5]
For 980, the last digit is 0. Therefore, it is divisible by 5 and the remainder is 0.
For 1111, the last digit is 1. Therefore, when 1111 is divided by 5, the remainder is 1. [∵ 1 < 5]
For 2345, the last digit is 5. Therefore, it is divisible by 5 and the remainder is 0.
(iii) We know that a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
For 78, the last digit is 8. Therefore it is divisible by 2 and the remainder is 0.
For 99, the last digit is 9. Therefore, when divided by 2, the remainder is 1.
For 173, the last digit is 3. Therefore, when divided by 2, the remainder is 1.
For 572, the last digit is 2. Therefore, when divided by 2, the remainder is 0.
For 980, the last digit is 0. Therefore, when divided by 2, the remainder is 0.
For 1111, the last digit is 1. Therefore, when divided by 2, the remainder is 1. [∵ 1 < 2]
For 2345, the last digit is 5. Therefore, when divided by 2, the remainder is 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:
If 14560 is divisible by 8 and 10, then it is also divisible by all the given numbers.
Only the last three digits matter when deciding if a given number is divisible by 8 as it is also divisible by 2 and 4.
If the unit place is 0 it is divisible by 10 and 5.
Question 6.
Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160?
Solution:
5600, 6000, 77622160
Question 7.
Write two numbers whose product is 10000. The two numbers should not have 0 as the unit digit.
Solution:
Prime factorisation of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.
So, the two numbers are 16 and 625 whose product is 10000.