Check below the NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers:
Content:
1. Book PDF
2. Exercise 3.1
3. Exercise 3.2
4. Exercise 3.3
5. Exercise 3.4
6. Exercise 3.5
7. Exercise 3.6
8. Exercise 3.7
Book PDF:1. Book PDF
2. Exercise 3.1
3. Exercise 3.2
4. Exercise 3.3
5. Exercise 3.4
6. Exercise 3.5
7. Exercise 3.6
8. Exercise 3.7
NCERT Books for class 6 Maths Chapter 3: Playing with Numbers in English Medium is given below to download in PDF form free. There are total 14 chapters in grade 6 Maths. All chapters are compulsory for the school test or terminal exams.The first chapter is Knowing Our Numbers and the last chapter is Practical Geometry. All the chapters are equally important for exams. Solutions of each chapter in English Medium or Hindi Medium is available on LEAF EDU91 website.
Exercise 3.1:
Write all the factors of the following numbers :
Q. 24
Ans. We have,
24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
24 = 6 x 4
Stop here, because 4 and 6 have occurred earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Q. 15
Ans. We have,
15 = 1 x 15
15 = 3 x 5
15 = 5 x 3
Stop here, because 3 and 5 have occurred earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.
Q. 21
Ans. We have,
21 = 1 x 21
21 = 3 x 7
21 = 7 x 3
Stop here, because 3 and 7 have occurred earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.
Q. 27
Ans. We have,
27 = 1 x 27
27 = 3 x 9
27 = 9 x 3
Stop here, because 3 and 9 have occurred earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.
Q. 12
Ans.We have,
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
12 = 4 x 3
Stop here, because 3 and 4 have occurred earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.
Q. 20
Ans. We have,
20 = 1 x 20
20 = 2 x10
20 = 4 x 5
20 = 5 x 4
Stop here, because 4 and 5 have occurred earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.
Q. 18
Ans. We have,
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
Stop here, because 3 and 6 have occurred earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.
Q. 23
Ans. We have,
23 = 1 x 23
23 = 23 x 1
Stop here, because 1 and 23 have occurred earlier.
Thus, all the factors of 23 are 1 and 23.
Q. 36
Ans. We have,
36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 1 x 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Q. 24
Ans. We have,
24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
24 = 6 x 4
Stop here, because 4 and 6 have occurred earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Q. 15
Ans. We have,
15 = 1 x 15
15 = 3 x 5
15 = 5 x 3
Stop here, because 3 and 5 have occurred earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.
Q. 21
Ans. We have,
21 = 1 x 21
21 = 3 x 7
21 = 7 x 3
Stop here, because 3 and 7 have occurred earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.
Q. 27
Ans. We have,
27 = 1 x 27
27 = 3 x 9
27 = 9 x 3
Stop here, because 3 and 9 have occurred earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.
Q. 12
Ans.We have,
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
12 = 4 x 3
Stop here, because 3 and 4 have occurred earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.
Q. 20
Ans. We have,
20 = 1 x 20
20 = 2 x10
20 = 4 x 5
20 = 5 x 4
Stop here, because 4 and 5 have occurred earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.
Q. 18
Ans. We have,
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
Stop here, because 3 and 6 have occurred earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.
Q. 23
Ans. We have,
23 = 1 x 23
23 = 23 x 1
Stop here, because 1 and 23 have occurred earlier.
Thus, all the factors of 23 are 1 and 23.
Q. 36
Ans. We have,
36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 1 x 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Write first five multiples of:
Q. 5
Ans. In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4=20
5 x 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.
Q. 8
Ans. In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4 = 32
8 x 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively.
Q. 9
Ans. In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.
Q. 5
Ans. In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4=20
5 x 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.
Q. 8
Ans. In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4 = 32
8 x 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively.
Q. 9
Ans. In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.
Q. Match the items in column 1 with the items in column 2.
Ans.
Ans.
| Column 1 | Column 2 | Explain |
|---|---|---|
| 35 | (c) Multiple of 70 | ∵ 35 x 2 = 70 |
| 15 | (d) Factor of 30 | ∵ 30 = 15 = 2 |
| 16 | (a) Multiple of 8 | ∵ 8 x 2 = 16 |
| 20 | (f) Factor of 20 | ∵ 20 = 20 = 1 |
| 25 | (e) Factor of 50 | ∵ 50 = 25 = 2 |
Q. Find all the multiples of 9 upto 100.
Ans. All the multiples of 9 upto 100 are
9 x 1, 9 x 2, 9 x 3, 9 x 4, 9 x 5, 9 x 6, 9 x 7, 9 x 8, 9 x 9, 9 x 10 and 9 x 11.
i. e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
Ans. All the multiples of 9 upto 100 are
9 x 1, 9 x 2, 9 x 3, 9 x 4, 9 x 5, 9 x 6, 9 x 7, 9 x 8, 9 x 9, 9 x 10 and 9 x 11.
i. e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
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Exercise 3.2:
What is the sum of any two-
(a) Odd numbers?
Ans. Sum of two odd numbers is even.
(b) Even numbers?
Ans. Sum of two even numbers is even.
(a) Odd numbers?
Ans. Sum of two odd numbers is even.
(b) Even numbers?
Ans. Sum of two even numbers is even.
State whether the following statements are True or False :
Q. The sum of three odd numbers is even.
Ans. False
Q. The sum of two odd numbers and one even number is even.
Ans. True
Q. The product of three odd numbers is odd.
Ans. True
Q. If an even number is divided by 2, the quotient is always odd.
Ans. False
Q. All prime numbers are odd.
Ans. False
Q. Prime numbers do not have any factors.
Ans. False
Q. Sum of two prime numbers is always even.
Ans. False
Q. 2 is the only even prime number.
Ans. True
Q. All even numbers are composite numbers.
Ans. False
Q. The product of two even numbers is always even.
Ans. True
Q. The sum of three odd numbers is even.
Ans. False
Q. The sum of two odd numbers and one even number is even.
Ans. True
Q. The product of three odd numbers is odd.
Ans. True
Q. If an even number is divided by 2, the quotient is always odd.
Ans. False
Q. All prime numbers are odd.
Ans. False
Q. Prime numbers do not have any factors.
Ans. False
Q. Sum of two prime numbers is always even.
Ans. False
Q. 2 is the only even prime number.
Ans. True
Q. All even numbers are composite numbers.
Ans. False
Q. The product of two even numbers is always even.
Ans. True
Q. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Ans. By the Sieve of Eratosthenes method find the prime numbers between 1 and 100. We find that these 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.
Out of these, a pair of prime numbers having same digits are 13, 31; 17, 71; 37, 73, 79, 97.
Ans. By the Sieve of Eratosthenes method find the prime numbers between 1 and 100. We find that these 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.
Out of these, a pair of prime numbers having same digits are 13, 31; 17, 71; 37, 73, 79, 97.
Q. Write down separately the prime and composite numbers less than 20.
Ans. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.
Ans. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.
Q. What is the greatest prime number between 1 and 10?
Ans. Prime numbers between 1 and 10 are 2, 5 and 7.
∴ Greatest prime number between 1 and 10 = 7
Ans. Prime numbers between 1 and 10 are 2, 5 and 7.
∴ Greatest prime number between 1 and 10 = 7
Express the following as the sum of two odd primes
Q. 44
Ans. 44 = 13 + 31
Q. 36
Ans. 36 = 5 + 31
Q. 24
Ans. 24 = 11 +13
Q. 18
Ans. 18 = 7+11
Note : In 1742, mathematician Goldbach had a conjecture (guess) for which he could not provide a proof. It may be stated as “Every even number greater than 4 can be expressed as the sum of two odd prime numbers”.
Q. 44
Ans. 44 = 13 + 31
Q. 36
Ans. 36 = 5 + 31
Q. 24
Ans. 24 = 11 +13
Q. 18
Ans. 18 = 7+11
Note : In 1742, mathematician Goldbach had a conjecture (guess) for which he could not provide a proof. It may be stated as “Every even number greater than 4 can be expressed as the sum of two odd prime numbers”.
Q. Give three pairs of prime numbers whose difference is 2.
Ans. Three pairs of prime number whose difference is 2 are 3, 5; 5, 7 and 11, 13.
Note : Two prime numbers are known as twin-primes if there is one composite number between them. In other words, two prime numbers whose difference is 2 are called twin-primes.
Ans. Three pairs of prime number whose difference is 2 are 3, 5; 5, 7 and 11, 13.
Note : Two prime numbers are known as twin-primes if there is one composite number between them. In other words, two prime numbers whose difference is 2 are called twin-primes.
Which of the following numbers are prime?
Q. 23
Ans. We find that 23 is not exactly divisible by any of the prime numbers 2, 3, 5, 7 and 11 (i.e., upto half of 23). So, it is a prime number.
Q. 51
Ans. We find that 51 is divisible by 3. So, it is not a prime number.
Q. 37
Ans. We find that 37 is not exactly divisible by any of the prime numbers 2, 3, 5, 7, 11, 13 and 17 (i.e., upto half of 37). So, it is a prime number.
Q. 26
Ans. We find that 26 is exactly divisible by 2 and 13. So, it is not a prime number.
Q. 23
Ans. We find that 23 is not exactly divisible by any of the prime numbers 2, 3, 5, 7 and 11 (i.e., upto half of 23). So, it is a prime number.
Q. 51
Ans. We find that 51 is divisible by 3. So, it is not a prime number.
Q. 37
Ans. We find that 37 is not exactly divisible by any of the prime numbers 2, 3, 5, 7, 11, 13 and 17 (i.e., upto half of 37). So, it is a prime number.
Q. 26
Ans. We find that 26 is exactly divisible by 2 and 13. So, it is not a prime number.
Q. Write seven consecutive composite numbers less-than 100 so that there is no prime number between them.
Ans. Seven consecutive composite numbers less than 100 so that there is no prime number between them are 90, 91, 92, 93, 94, 95 and 96.
Ans. Seven consecutive composite numbers less than 100 so that there is no prime number between them are 90, 91, 92, 93, 94, 95 and 96.
Express each of the following numbers as the sum of three odd primes:
Q. 21
Ans. 21 = 3 + 5 + 13
Q. 31
Ans. 31 = 3 + 5 + 23
Q. 53
Ans. 53 = 3 + 19 + 31
Q. 61
Ans. 61 = 3 + 11 + 47
Q. 21
Ans. 21 = 3 + 5 + 13
Q. 31
Ans. 31 = 3 + 5 + 23
Q. 53
Ans. 53 = 3 + 19 + 31
Q. 61
Ans. 61 = 3 + 11 + 47
Q. Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
Ans. Prime numbers below 20 are 2, 3, 5, 7, 13, 17 and 19.
Possible sum of pairs of these numbers :
Clearly, five pairs of prime numbers whose sum is divisible by 5 are 2, 3; 2, 13; 3, 7; 3,17 and 7, 13.
Ans. Prime numbers below 20 are 2, 3, 5, 7, 13, 17 and 19.
Possible sum of pairs of these numbers :
Fill in the blanks :
Q.
Ans.
Q.
- A number which has only two factors is called a
- A number which has more than two factors is called a
- 1 is neither nor
- The smallest prime number is
- The smallest composite number is
- The smallest even number is
Ans.
- Prime
- composite
- prime, composite
- 2
- 4
- 2
Exercise 3.3:
Q. Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 4; by 11 (say yes or no) :
Ans.
Ans.
Q. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8 :
(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150
Ans.
We know that a number is divisible by 4, if the number formed by its digits in ten’s and unit’s place is divisible by 4.
(a) In 572, 72 is divisible by 4. So, 572 is divisible by 4.
(b) In 726352, 52 is divisible by 4. So, it is divisible by 4.
(c) In 5500, 00 is divisible by 4. So, it is divisible by 4.
(d) In 6000, 00 is divisible by 4. So, it is divisible by 4.
(e) In 12159, 59 is not divisible by 4. So, it is not divisible by 4.
(f) In 14560, 60 is divisible by 4. So, it is divisible by 4.
(g) In 21084, 84 is divisible by 4. So, it is divisible by 4.
(h) In 31795072, 72 is divisible by 4. So, it is divisible by 4.
(i) In 1700,00 is divisible by 4. So, it is divisible by 4.
(j) In 2150, 50 is not divisible by 4. So, it is not divisible by 4.
Also, we know that a number is divisible by 8, if the number formed by its hundred’s, ten’s and unit’s places is divisible by 8.
(a) 572 is not divisible by 8.
(b) In 726352, 352 is divisible by 8. So, it divisible by 8.
(c) In 5500, 500 is not divisible by 8. So, it is not divisible by 8.
(d) In 6000, 000 is divisible by 8. So, it is divisible by 8.
(e) In 12159, 159 is not divisible by 8. So, it is not divisible by 8.
(f) In 14560, 560 is divisible by 8. So, it is divisible by 8.
(g) In 21084, 084 is not divisible by 8. So, it is not divisible by 8.
(h) In 31795072, 072 is divisible by 8. So, it is divisible by 8.
(i) In 1700, 700 is not divisible by 8. So, it is not divisible by 8.
(j) In 2150, 150 is not divisible by 8. So, it is not divisible by 8.
(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150
Ans.
We know that a number is divisible by 4, if the number formed by its digits in ten’s and unit’s place is divisible by 4.
(a) In 572, 72 is divisible by 4. So, 572 is divisible by 4.
(b) In 726352, 52 is divisible by 4. So, it is divisible by 4.
(c) In 5500, 00 is divisible by 4. So, it is divisible by 4.
(d) In 6000, 00 is divisible by 4. So, it is divisible by 4.
(e) In 12159, 59 is not divisible by 4. So, it is not divisible by 4.
(f) In 14560, 60 is divisible by 4. So, it is divisible by 4.
(g) In 21084, 84 is divisible by 4. So, it is divisible by 4.
(h) In 31795072, 72 is divisible by 4. So, it is divisible by 4.
(i) In 1700,00 is divisible by 4. So, it is divisible by 4.
(j) In 2150, 50 is not divisible by 4. So, it is not divisible by 4.
Also, we know that a number is divisible by 8, if the number formed by its hundred’s, ten’s and unit’s places is divisible by 8.
(a) 572 is not divisible by 8.
(b) In 726352, 352 is divisible by 8. So, it divisible by 8.
(c) In 5500, 500 is not divisible by 8. So, it is not divisible by 8.
(d) In 6000, 000 is divisible by 8. So, it is divisible by 8.
(e) In 12159, 159 is not divisible by 8. So, it is not divisible by 8.
(f) In 14560, 560 is divisible by 8. So, it is divisible by 8.
(g) In 21084, 084 is not divisible by 8. So, it is not divisible by 8.
(h) In 31795072, 072 is divisible by 8. So, it is divisible by 8.
(i) In 1700, 700 is not divisible by 8. So, it is not divisible by 8.
(j) In 2150, 150 is not divisible by 8. So, it is not divisible by 8.
Using divisibility tests, determine which of the following numbers are divisible by 6 :
a. 297144
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 297144
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.
∴ 297144 is divisible by 6.
b. 1258
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number =1258
Its unit’s digit is 8. So, it is divisible by 2.
Sum of its digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.
∴ 1258 is not divisible by 6.
c. 4335
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 4335 .
Its unit’s digit is 5. So, it is not divisible by 2.
∴ 4335 is also not divisible by 6.
d. 61233
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 61233
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 61233 is also not divisible by 6.
e. 901352
Ans.We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 901352
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.
∴ 901352 is not divisible by 6.
f. 438750
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 438750
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.
∴ 438750 is divisible by 6.
g. 1790184
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 1790184
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.
∴ 1790184 is divisible by 6.
h. 12583
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 12583 .
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 12583 is not divisible by 6.
i. 639210
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 639210
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.
∴ 639210 is divisible by 6.
j. 17852
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 17852
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.
∴ 17852 is not divisible by 6.
a. 297144
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 297144
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.
∴ 297144 is divisible by 6.
b. 1258
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number =1258
Its unit’s digit is 8. So, it is divisible by 2.
Sum of its digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.
∴ 1258 is not divisible by 6.
c. 4335
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 4335 .
Its unit’s digit is 5. So, it is not divisible by 2.
∴ 4335 is also not divisible by 6.
d. 61233
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 61233
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 61233 is also not divisible by 6.
e. 901352
Ans.We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 901352
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.
∴ 901352 is not divisible by 6.
f. 438750
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 438750
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.
∴ 438750 is divisible by 6.
g. 1790184
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 1790184
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.
∴ 1790184 is divisible by 6.
h. 12583
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 12583 .
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 12583 is not divisible by 6.
i. 639210
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 639210
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.
∴ 639210 is divisible by 6.
j. 17852
Ans. We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
Given number = 17852
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.
∴ 17852 is not divisible by 6.
Using divisibility tests, determine which of the following numbers are divisible by 11:
a. 5445
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 5445
Sum of its digits at odd places = 5 + 4 = 9
Sum of its digit at even places = 4 + 5 = 9
Difference of these two sums = 9 – 9 = 0
∴ 5445 is divisible by 11.
b. 10824
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 10824
Sum of its digits at odd places = 4+ 8 + 1 = 13
Sum of its digits at even places =2 + 0 =2
Difference of these two sums =13 – 2 = 11, which is a multiple of 11.
∴ 10824 is divisible by 11.
c. 7138965
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 7138965
Sum of its digits at odd places = 5+ 9+ 3 + 7= 24
Sum of its digits at even places = 6+ 8 + 1 = 15
Difference of these two sums = 24 – 15 = 9, which is not a multiple of 11.
∴ 7138965 is not divisible by 11.
d. 70169308
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 70169308
Sum of its digits at odd places = 8 + 3 + 6 + 0=17
Sum of its digits at even places = 0 + 9 + 1 + 7 = 17
Difference of these two sums =17 – 17 = 0
∴ 70169308 is divisible by 11.
e. 10000001
Ans.We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 10000001
Sum of its digits at odd places = 1 + 0 + 0 + 0 = 1
Sum of its digits at even places = 0 + 0 + 0 + 1 = 1
Difference of these two sums = 1 – 1 = 0
∴ 10000001 is divisible by 11.
f. 901153
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 901153
Sum of its digits at odd places = 3 + 1 + 0 = 4
Sum of its digits at even places = 5 + 1 + 9=15
Difference of these two sums =15 – 4 = 11,
which is a multiple of 11.
∴ 901153 is divisible by 11.
a. 5445
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 5445
Sum of its digits at odd places = 5 + 4 = 9
Sum of its digit at even places = 4 + 5 = 9
Difference of these two sums = 9 – 9 = 0
∴ 5445 is divisible by 11.
b. 10824
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 10824
Sum of its digits at odd places = 4+ 8 + 1 = 13
Sum of its digits at even places =2 + 0 =2
Difference of these two sums =13 – 2 = 11, which is a multiple of 11.
∴ 10824 is divisible by 11.
c. 7138965
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 7138965
Sum of its digits at odd places = 5+ 9+ 3 + 7= 24
Sum of its digits at even places = 6+ 8 + 1 = 15
Difference of these two sums = 24 – 15 = 9, which is not a multiple of 11.
∴ 7138965 is not divisible by 11.
d. 70169308
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 70169308
Sum of its digits at odd places = 8 + 3 + 6 + 0=17
Sum of its digits at even places = 0 + 9 + 1 + 7 = 17
Difference of these two sums =17 – 17 = 0
∴ 70169308 is divisible by 11.
e. 10000001
Ans.We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 10000001
Sum of its digits at odd places = 1 + 0 + 0 + 0 = 1
Sum of its digits at even places = 0 + 0 + 0 + 1 = 1
Difference of these two sums = 1 – 1 = 0
∴ 10000001 is divisible by 11.
f. 901153
Ans. We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Given number = 901153
Sum of its digits at odd places = 3 + 1 + 0 = 4
Sum of its digits at even places = 5 + 1 + 9=15
Difference of these two sums =15 – 4 = 11,
which is a multiple of 11.
∴ 901153 is divisible by 11.
Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :
a. … 6724
Ans. We know that a number divisible by 3, if the sum of its digits is divisible by 3.
For … 6724, we have 6 + 7 + 2 + 4 =19, we add 2 to 19, the resulting number 21 will be divisible by 3.
∴ The required smallest digit is 2.
Again, if we add 8 to 19, the resulting number 27 will be divisible by 3. .’. The required largest digit is 8.
b. 4765 … 2
Ans. We know that a number divisible by 3, if the sum of its digits is divisible by 3.
For 4765 … 2, we have 4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.
Hence the required smallest digit is 0.
Again, if we add 9 to 24, the resulting number 33 will be divisible by 3.
∴ The required largest digit is 9.
a. … 6724
Ans. We know that a number divisible by 3, if the sum of its digits is divisible by 3.
For … 6724, we have 6 + 7 + 2 + 4 =19, we add 2 to 19, the resulting number 21 will be divisible by 3.
∴ The required smallest digit is 2.
Again, if we add 8 to 19, the resulting number 27 will be divisible by 3. .’. The required largest digit is 8.
b. 4765 … 2
Ans. We know that a number divisible by 3, if the sum of its digits is divisible by 3.
For 4765 … 2, we have 4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.
Hence the required smallest digit is 0.
Again, if we add 9 to 24, the resulting number 33 will be divisible by 3.
∴ The required largest digit is 9.
Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
a. 92 … 389
Ans. We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
For 92 … 389, sum of the digits at odd places and sum of digits at even places
= 9 + 3 + 2 = 14
= 8 + required digit + 9
= required digit+ 17
Difference between these sums
= required digit + 17 – 14
= required digit + 3
For (required digit + 3) to become 11, we must have the required digit as 8 (∵ 3+ 8 gives 11).
Hence, the required smallest digit = 8 .
b. 8 … 9484
Ans. We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
For 8 … 9484, sum of the digits at odd places
= 4 + 4 + required digit
= 8 + required digit
and sum of digits at even places
= 8 + 9 + 8 = 25
Difference between these sums
= 25 – (8 + required digit)
= 17- required digit
For (17 — required digit) to become 11 we must have the required digit as 6 (∵ 17-6 =11).
Hence the required smallest digit = 6
a. 92 … 389
Ans. We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
For 92 … 389, sum of the digits at odd places and sum of digits at even places
= 9 + 3 + 2 = 14
= 8 + required digit + 9
= required digit+ 17
Difference between these sums
= required digit + 17 – 14
= required digit + 3
For (required digit + 3) to become 11, we must have the required digit as 8 (∵ 3+ 8 gives 11).
Hence, the required smallest digit = 8 .
b. 8 … 9484
Ans. We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
For 8 … 9484, sum of the digits at odd places
= 4 + 4 + required digit
= 8 + required digit
and sum of digits at even places
= 8 + 9 + 8 = 25
Difference between these sums
= 25 – (8 + required digit)
= 17- required digit
For (17 — required digit) to become 11 we must have the required digit as 6 (∵ 17-6 =11).
Hence the required smallest digit = 6
Exercise 3.4:
