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 No. 108)
Question 1.
At what number is ‘idli-vada’ said for the 10th time?
Solution:
When the players arrive at multiples of 3, players 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’
So, by the above information multiples of 3 and 5 are 15, 30, 45, and so on.
Idli Vada is said for the 10th time at 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) Idli is said for multiples of 3. Between 1 and 90 the multiples of 3 are 3, 6, 9, 12, 15, 18, …….. 90.
There are 30 such numbers.
Hence the children would say idli 30 times.
(b) Vada is said for multiples of 5. Between 1 and 90 the multiples of 5 are 5, 10,15, 20, 25,…
There are 18 such numbers.
(c) Idli-Vada is said for multiples of both 3 and 5, which is multiple of 15. Between 1 and 90, there are 15, 30,45,60,75,90. There are 6 such numbers.
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?
Solution:
In the given figure there are two numbers 15 and 30 which are multiple of 3 and 5 both, When we draw till 60.
There are 20 numbers which are multiples of 3 – 3, 6, 9, 12, 15, 18,…..,57, 60,
And 12 numbers which are multiple of 5 are 5, 10, 15, 20,25,……,55, 60
And 4 common multiples of 3 and 5 are 15, 30, 45, and 60.
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:
Here, multiples df 40 are 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440
Hence 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 l more than the other.
Solution:
(a) 7 is the common factors of 7, 14, 21,28, 35, which are less than 40. And there is one number which have digit sum of 8, is 35 = (3 + 5) = 8.
So, I am 35.
(b) Common factors of 3 and 5 are 15, 30, 45, 60, 75,90, (which are less than 100). And there is one number which one of digit is 1 more than the other that is 45. So, I am 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:
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:
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:
Multiples of 7 are 7,14, 21, 28, 35,42, 49, 56,…
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56,…
Multiples of both 7 and 8 are 56, 112, 168,…
So, the first time anybody will say ‘idli-vada’ after the number 50 is 56.
Thus, the two numbers which are assigned ‘idli’ and ‘vada’ are 7 and 8, respectively.
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 are 1, 2,4, 7,14 and 28.
Factors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70.
Common factors of 28 and 70 are 1, 2, 7 and 14.
Thus, the jump sizes of 1, 2, 7 and 14 will land on both 28 and 70.
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:
Multiples of 8 are 8, 16, 24, 32,…..
Multiples of 12 are 12, 24, 36, 48,……
Common multiples of 8 and 12 are 24, 48, 72,…
Therefore, the numbers at blank spaces are 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:
Take the LCM of 1, 2, 3, 4, 5, 6, 8, 9, and 10 (excluding 7) which is 360.
So, it is the smallest number which is a multiple of all except 7.
Question 10.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Solution:
To find the shiallesf number that is a multiple of all
numbers from 1 to 10, we need to determine LCM of the numbers from 1 to 10.
Prime factorisation of numbers from 1 to 10.
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 (1 to 10) = 23 × 32 × 5 × 7
= 2520
Thus, the required smallest number is 2520.
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 even prime number. Since 2 is the only even number that meets the criteria of a prime number (its only divisors are 1 and 2), it is the only even prime number. All other even numbers are divisible by 2 and at least one other number, so they are not prime.
Question 2.
Look at the list of primes till 106. What is the smallest difference between two successive primes? What is the largest difference?
Solution:
To find the smallest difference between two successive prime numbers up to 100, let’s list the prime numbers in that range and calculate the differences between each pair: Prime numbers up to 100: 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
Differences between successive primes:
- 3 – 2 = 1
- 5 – 3 = 2
- 7 – 5 = 2
- 11 – 7 = 4
- 13 – 11 = 2
- 17 – 13 = 4
- 19 – 17 = 2
- 23 – 19 = 4
- 29 – 23 = 6
- 31 – 29 = 2
- 37 – 31 = 6
- 41 – 37 = 4
- 43 – 41 = 2
- 47 – 43 = 4
- 53 – 47 = 6
- 59 – 53 = 6
- 61 – 59 = 2
- 67 – 61 = 6
- 71 – 67 = 4
- 73 – 71 = 2
- 79 – 73 = 6
- 83 – 79 = 4
- 89 – 83 = 6
- 97 – 89 = 8
The smallest difference between two successive primes up to 100 is 1 (between the primes 2 and 3).
The largest difference between two successive primes up to 100 is 8, which occurs between the primes 89 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, 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:
Prime numbers less than 20 are
2, 3, 5,7,11,13,17 and 19.
Now, 2 + 3 = 5 (multiple of 5)
2 + 13 = 15 (multiple of 5)
7 +13 = 20 (multiple of 20)
Thus, three pairs 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 having same digits upto 100 are 17 and 71 (both have digits 1 and 7) 37 and 73 (both have digits 3 and 7)
Question 7.
Find seven consecutive composite numbers between 1 and 100.
Solution:
The 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 l and 100.
Solution:
The twin primes between 1 and 100 other than 3 and 5,17 and 19 are as follows: (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.
A prime number must end in 1, 3, 7, or 9 (except for the number 2) because any number ending in 0, 2, 4, 6 or 8 is divisible by 2.
Thus, there is no prime number whose units digit is 4.
(b) False.
A product of prime numbers is only prime if it involves exactly one prime number. When you multiply two or more prime numbers together, the result is always a composite number, not a prime. As this number has 2 factors now.
(c) False.
Prime numbers have exactly two factors 1 and itself.
(d) False.
The number 2 is an even number, but it is not composite. As it is a prime number.
(e) True.
For every prime number greater than 2, the 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:
Prime factorisation of 45, 60, 91, 105 and 330 are
45 = 3 × 3 × 5
60 = 2 × 2 × 3 × 5
91 = 7 × 13
105 = 3 × 5 × 7
and 330 = 2 × 3 × 5 × 11
The number 105 is product of exactly three distinct primes.
Question 11.
How many three-digit prime numbers can you make using each of 2, 4, and 5 once?
Solution:
There is no prime number that can be made using the digits 2, 4, and 5 because if the unit place is 2 or 4, then it is divisible by 2 and if the unit place is 5, then it is divisible by 5.
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 such examples are
2 × 2 + 1 = 5 (prime number)
2 × 5 + 1 = 11 (prime number),
2 × 11 + 1 = 23 (prime number),
2 × 23 + 1 = 47 (prime number)
and 2 × 29 + 1 = 59 (prime number)
5.3 Co-prime Numbers for Safekeeping Treasures Figure it Out (Page No. 120)
Question 1.
Find the prime factorizations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.
Solution:
Prime factor of
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
100 = 2 × 2 × 2 × 5 × 5 × 5
Question 2.
The prime factorization of a number has one 2, two 3s, and one 11. What is the number?
Solution:
The number which has one 2, two 3s, and one 11 is 2 × 3 × 3 × 11 = 198
Question 3.
Find three prime numbers, all less than 30, whose product is 1955.
Solution:
The prime factors of 1955 are 5 × 17 × 19.
All numbers are less than 30.
Question 4.
Find the prime factorization 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
Combined prime factorization of 56 × 25 = 2 × 2 × 2 × 7 × 5 × 5
(b) Prime factors of 108 = 2 × 2× 2 × 3 × 3
Prime factors of 75 = 3 × 5 × 5
Combined prime factorization 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
Combined prime factorization of 1000 × 81 = 2 × 2 × 2 × 5 × 5 × 5 × 3 ×3 × 3 × 3
Question 5.
What is the smallest number whose prime factorization has:
(a) three different prime numbers?
(b) four different prime numbers?
Solution:
(a) The smallest prime numbers are 2, 3, and 5. To find the smallest number with these primes as factors, multiply them together:
2 × 3 × 5 = 30
So, the smallest number whose prime factorization has three different prime numbers is 30.
(b) The smallest four prime numbers are 2, 3, 5, and 7. To find the smallest number with these primes as factors, multiply them together:
2 × 3 × 5 × 7 = 210
Thus, the smallest number whose prime factorization has four different prime numbers is 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:
(i) No, 30 and 45 are not coprime.
Prime factorisation of 30 and 45 are
30 = 2 × 3 × 5
45 = 3 × 3 × 5
We see that 3 and 5 are prime factors of 30 and 45 both.
Therefore, 30 and 45 are not coprime.
(ii) Yes, 57 and 85 are coprime.
Prime factorisation of 57 and 85 are
57 = 3 × 19
85 = 5 × 17
We see that there are no common primes that divides both 57 and 85.
Therefore, 57 and 85 are coprime.
(iii) No, 121 and 1331 are not coprime.
Prime factorisation of 121 and 1331 are
121 = 11 × 11
1331 = 11 × 11 × 11
We see that 11 is the prime factor of both 121 and 1331.
Therefore, 121 and 1331 are not coprime.
(iv) Yes, 343 and 216 are coprime.
Prime factorisation of 343 and 216 are
343 = 7 × 7 × 7
216 = 2 × 2 × 2 × 3 × 3 × 3
We see that there are no common primes that divides both 343 and 216.
Therefore, 343 and 216 are coprime.
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:
Prime factorisation of two numbers are 2 × 3 × 7 and 3 × 7 × 11.
Both the numbers have a common factor 3.
Therefore, they are not co-prime.
Since, the prime factors of one number are not the prime factors of another number, therefore one of them does not divide the other.
Question 4.
Guna says, “Any two prime numbers are co-prime”. Is he right?
Solution:
Yes, any two distinct prime numbers are coprime. This is because prime numbers have no positive integer divisors other than 1 and themselves.
Therefore, any two prime numbers are coprime.
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 born till now, which years were leap years?
(b) From the year 2024 till 2099, how many leap years are there?
Solution:
(a) If you were born in *2006, the leap years would be 2008, 2012, 2016, 2020 and 2024.
(b) From the year 2024 till 2099, there are 19 leap years – 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092 and 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) (Sometimes true)
e.g. Let the two even numbers he 2 and 8, then 2+8=10, which is not a multiple of 4.
Again, let two even numbers be 8 and 4, then 8 +4 = 12, which is a multipe of 4.
(b) (Sometimes true)
e.g. Let the two odd numbers be 3 and 7, then 3 + 7 = 10, which is not a multiple of 4.
Again, let two odd numbers 3 and 5, then 3 + 5=8, which is a multiple of 4.
Question 4.
Find the remainder obtained when each of the following numbers is divided by (i) 10, (ii) 5, (iii) 2.
78, 99, 173, 572, 980, 1111, 2345
Solution:
When 78 is divided by 10 then the remainder is 8
When 78 is divided by 5 then the remainder is 3
When 78 is divided by 2 then the remainder is 0
When 99 is divided by 10 then the remainder is 9
When 99 is divided by 5 then the remainder is 4
When 99 is divided by 2 then the remainder is 1
When 173 is divided by 10 then the remainder is 3
When 173 is divided by 5 then the remainder is 3
When 173 is divided by 2 then the remainder is 1
When 572 is divided by 10 then the remainder is 2
When 572 is divided by 5 then the remainder is 2
When 572 is divided by 2 then the remainder is 0
When 980 is divided by 10 then the remainder is 0
When 980 is divided by 5 then the remainder is 0
When 980 is divided by 2 then the remainder is 0
When 1111 is divided by 10 then the remainder is 1
When 1111 is divided by 5 then the remainder is 1
When 1111 is divided by 2 then the remainder is 1
When 2345 is divided by 10 then the remainder is 5
When 2345 is divided by 5 then the remainder is 0
When 2345 is divided by 2 then 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:
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 unit digit.
Solution:
Two number whose product is 10000 are 16 and 625.
i. e. 16 × 625 = 10000.
Both 16 and 625 do not have 0 as the unit digit.