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:
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:
(i) How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)?
(ii) How many times would the children say ‘vada’ (including the times they coy ‘idli-vada’)?
(iii) How many times would the children say ‘idli-vada’?
Solution:
(i) The multiples of 3 from 1 till 90 are 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 and 90.
Thus, the number of times, the children would say ‘idli’ is 30.
(ii) The multiples of 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55,60, 65, 70, 75, 80, 85 and 90.
Thus, the number of times, the children would say ‘vada’ is 18.
(iii) The multiples of 15 are 15, 30,45,60, 75 and 90. Thus, the number of times, the children would say ‘idli-vada’ is 6.
Question 3.
What if the game was played till 900? How would your answers change?
Solution:
Number of multiples of 3 from 1 till 900 = \(\frac{900}{3}\) = 300
Number of multiples of 5 from 1 till 900 = \(\frac{900}{5}\) = 180
Number of multiples of 15 from 1 till 900 = \(\frac{900}{15}\) = 60
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.
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:
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 numbers less than 40 whose one of the factor is 7 are 7,14, 21, 28, 35. Out of these numbers the sum of digits of 35 is 8.
Hence, the number less than 40 whose one of the factors is 7 and sum of digits equals to 8 is 35.
(b) The number less than 100 whose two factors 3 and 5 are 15, 30, 45,60, 75, 90. Out of these numbers, the number 15 has one of digits is 1 more than the other.
Hencd, the number less than 100 whose two factors are 3 and 5 and one of digits is 1 more than the other is 15.
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:
Factors of 6 are 1,2 and 3.
⇒ 1 + 2 + 3 + 6 = 12
Since, 12 = 2 × 6
∴ Sum of factors is equal to twice the number.
Thus, 6 is a perfect number between 1 and 10.
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 are 1,2,4, 5, 10 and 20.
Factors of 28 are 1, 2,4, 7, 14 and 28.
Common factors of 20 and 28 are 1, 2 and 4.
(b) Factors of 35 are 1, 5, 7 and 35.
Factors of 50 are 1, 2, 5, 10, 25 and 50.
Common factors of 35 and 50 are 1 and 5.
(c) 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) 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:
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 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.
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:
Prime numbers upto 100 are 2, 3, 5, 7,11,13,17,19, 23, 29, 31, 37,41,43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97
Difference 3 – 2 = 1 and 97 – 89 = 8
Smallest difference: The smallest difference between successive prime is 1 (i.e. between 2 and 3).
Largest difference: The largest difference between successive prime is 8 (i.e. between 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, the number of primes occurring in every row are not equal.
The decade 91 to 100 has least number of primes i.e. 1.
The decades having most number of primes are 1 to 10 and 11 to 20 (4 primes)
Question 4.
Which of the following numbers are prime? 23, 51, 37, 26
Solution:
23 = 1 × 23, 23 = 23 × 1
Since, 23 has only two factors i.e. 1 and 23.
Therefore, 23 is a prime number.
51 = 1 × 51, 51 = 3 × 17
Since, 51 has four factors i.e. 1, 3,17 and 51.
Therefore, 51 is not a prime number.
37 = 1 × 37, 37 = 37 × 1
Since, 37 has only two factors i.e. 1 and 37.
Therefore, 37 is a prime number.
26 = 1 × 26, 26 = 2 × 13
Since, 26 has four factor i.e. 1,2,13 and 26.
Therefore, 26 is not a prime number.
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:
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:
Twin primes between 1 and 100 are
(3, 5) → 5 – 3 = 2
(5, 7) → 7 – 5 = 2
(11, 13) → 13 – 11 = 2
(17, 19) → 19 – 17 = 2
(29, 31) → 31 – 29 = 2
(41, 43) → 43 – 41 = 2
(59, 61) → 61 – 59 = 2
(71, 73) → 73 – 71 = 2
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)
Any number ending in 4 is divisible by 2, making it an even number.
Except for the number 2, no even number is prime.
(b) (False)
No, a product of primes cannot be a prime.
e.g. 2 and 3 are prime numbers but 2×3 = 6, which is not a prime number.
When you multiply two or more primes, you get a composite number, which has divisors other than just 1 and itself.
(c) (False)
Prime numbers have factors but only two
i. e. 1 and the number itself.
e.g. Prime number 5 has only two factors 1 and 5.
(d) (False)
No, all even numbers are not composite numbers, e.g. 2 is an even prime number.
(e) (True)
2 is a prime followed by a prime number 3.
For every other prime, the next number is composite, e.g. 5 followed by 6, which is composite.
11 followed by 12, which is composite.
13 followed by 14, which 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:
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:
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 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:
(a) Given number is 56 × 25.
Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 25 = 5 × 5
∴ Prime factorisation of 56 × 25
= 2 × 2 × 2 × 7 × 5 × 5 = 2 × 2 × 2 × 5 × 5 × 7
(b) Given number is 108 × 75.
Prime factorisation of 108 = 2 × 2 × 3 × 3 × 3
Prime factorisation of 75 = 3 × 5 × 5
∴ Prime factorisation of 108 × 75
= 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
(c) Given number is 1000 × 81.
Prime factorisation of 1000 = 2 × 2 × 2 × 5 × 5 × 5
Prime factorisation 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 three prime numbers are 2, 3 and 5.
Thus, the smallest number with exactly three different prime factors = 2 × 3 × 5 = 30
(b) The smallest four prime numbers are 2,3,5 and 7.
Thus, the smallest number with exactly four different prime factors = 2 × 3 × 5 × 7 = 210
5.4 Prime Factorisation Figure it Out (Page No. 122)
Question 1.
Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer.
(a) 30 and 45
(b) 57 and 85
(c) 121 and 1331
(d) 343 and 216
Solution:
(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:
(i) Prime factorisation of 225 and 27 are 225 = 3 × 3 × 5 × 5
27 = 3 × 3 × 3
All prime factors of 27 are also prime factors of 225. But the prime factorisation of 12 is not included in the prime factorisation of 225.
This is because 3 occurs thrice in the prime factorisation of 27 but twice in the prime factorisation of 225. This means that 225 is not divisible by 27.
(ii) Prime factorisation of 96 and 24 are
96 = 2 × 2 × 2 × 2 × 2 × 3,
24 = 2 × 2 × 2 × 3
Since, we can multiply in any order,
96 = 24 × 4
Therefore, 96 is divisible by 24.
(iii) Prime factorisation of 343 and 17 are
343 = 7 × 7 × 7 and 17 = 17
The prime factorisation of 343 is 7 × 7 × 7 and 17 is a prime number with no common factor with 343.
Therefore, 343 is not divisible by 17.
(iv) Prime factorisation of 999 and 99 are
999 = 3 × 3 × 111,
99 = 3 × 3 × 11
Prime factorisation of 99 is not included in the prime factorisation of 999.
Therefore, 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 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:
The smallest 4-digit palindrome is 1001.
Checking divisibility by 4 1001 (not divisible by 4)
1111 (not divisible by 4) 1221 (not divisible by 4)
1331 (not divisible by 4)
1441 (not divisible by 4)
1551 (not divisible by 4)
1661 (not divisible by 4)
1771 (not divisible by 4)
1881 (not divisible by 4)
1991 (not divisible by 4)
2002 (not divisible by 4)
2112 (divisible by 4)
Therefore, the smallest 4-digit palindromic number divisible by 4 is 2112.
The largest 4-digit palindrome is 9999.
Checking divisibility by 4 9999 (not divisible by 4)
9889 (not divisible by 4)
9779 (not divisible by 4)
9669 (not divisible by 4)
9559 (not divisible by 4)
9449 (not divisible by 4)
9339 (not divisible by 4)
9229 (not divisible by 4)
9119 (not divisible by 4)
9009 (not divisible by 4)
8998 (not divisible by 4)
8888 (divisible by 4)
Therefore, the largest 4-digit palindromic number divisible by 4 is 8888.
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 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:
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:
The last digit of 14560 is 0, therefore it is divisible by 5 and 10 both. Thus, 5 and 10 could be those two numbers.
Question 7.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Solution:
For 572
The last digit of 572 is 2, therefore it is divisible by 2.
The number formed by last two digits is divisible by 4,
therefore 572 is divisible by 4.
The last digit of 572 is not 0 or 5, therefore it is not divisible by 5.
The number 572 is not divisible by 8.
The last digit of 572 is not 0, therefore it is not divisible by 10.
For 2352
The last digit of 2352 is 2, therefore 2352 is divisible by 2.
The number formed by last two digits is divisible by 4, therefore 2352 is divisible by 4.
The last digit is not 0 or 5, therefore it is not divisible by 5.
The number formed by last 3-digits is divisible by 8, therefore it is divisible by 8.
The last digit of 2352 is not 0, therefore it is not divisible by 10.
For 5600
The last digit of 5600 is 0, therefore it is divisible by 2, 5 and 10.
The number formed by last two digits of 5600 is divisible by 4, therefore it is divisible by 4.
The number formed by last 3 digits is divisible by 8, therefore it is divisible by 8.
For 6000
The last digit of 6000 is 0, therefore it is divisible by 2, 5, and 10.
The number formed by last two-digits is divisible by 4, therefore it is divisible by 4.
The number formed by last three-digits is divisible by 8, therefore it is divisible by 8.
For 77622160
The last digit of 77622160 is 0, therefore it is divisible by 2, 5 and 10.
The number formed by last two digits of 77622160 is divisible by 4, therefore it is divisible by 4.
The number formed by last three digits of 77622160 is divisible by 8, therefore it is divisible by 8.
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.