ML Aggarwal Playing With Numbers Exe-4.2 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-4.2 Questions for Playing With Numbers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Playing With Numbers Exe-4.2 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-4 Playing With Numbers Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-4.2 Questions Edition 2023-2024

### Playing With Numbers Exe-4.2

ML Aggarwal Class 6 ICSE Maths Solutions

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#### Question 1. Which of the following numbers are divisible by 5 or by 10:

(i) 3725
(ii) 48970
(iii) 56823
(iv) 760035
(v) 7893217
(vi) 4500010

(i) 3725 : divisible by 5 as last digit is 5.

(ii) 48970 : divisible by 5 and 10 as last digit is 0.

(iii) 56823 : divisible by 5 and neither by 10  as last digit is 3.

(iv) 760035 : divisible by 5 as last digit is 5.

(v) 7893217 : not divisible by 5 and neither 10 as last digit is 7.

(vi) 4500010 : divisible by both 5 and 10 as last digit is 0.

#### Question 2. Which of the following numbers are divisible by 2, 4 or 8:

(i) 54014
(ii) 723840
(iii) 6531088
(iv) 75689604
(v) 786235
(vi) 5321048

(i) 54014

Last digit is 4 ,

hence it is divisible by 2 but not by 4 and 8.

(ii) 723840

This number is divisible by 8 ,

hence, it should get divide by all factors

(iii) 6531088

This number is divisible by 8 ,

hence, it should get divide by all factories 2 ,4 and 8

(iv) 75689604

This number is divisible by 4 and not by 8.

if it is divisible by 4 then it should also get divisible by its factors also ex. 2

(v) 786235

Last digit is 5 ,

The number is even.

hence, it s not divisible by 2, 4 and 8.

(vi) 5321048

This number is divisible by 8 ,

#### Question 3. Which of the following numbers are divisible by 3 or 9:

(i) 7341
(ii) 59031
(iii) 12345678
(iv) 560319
(v) 720634
(vi) 3721509

A number is divisible by 3 if the sum of its digit is divisible by 3 or 9.

(i) 7341 = 7+3+4+1 = 15 : this number is divisible by 3
(ii) 59031 = 5+9+0+3+1 = 18 : this number is divisible by 3,9
(iii) 12345678 = 1+2+3+4+5+6+7+8 = 36 : this number is divisible by 3,9
(iv) 560319 = 5+6+0+3+1+9 = 24 : this number is divisible by 3
(v) 720634 = 7+2+0+6+3+4 = 22 : this number is divisible by 3,9
(vi) 3721509 = 3+7+2+1+5+0+9 = 27 : this number is divisible by 3,9

### Playing With Numbers Exe-4.2

ML Aggarwal Class 6 ICSE Maths Solutions

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#### Question 4. Examine the following numbers for divisibility by 11:

(i) 10428
(ii) 70169803
(iii) 7136985

(i) 10428 = 1 + 4 + 8 = 13

and  0 + 2 = 2

Difference = 13-2 = 11

divisible by 11

(ii) 70169803 =  7 + 1 + 9 + 0 = 17

and 0 + 6 + 8 + 3 = 17

Difference = 17-17 = 0

divisible by 11

(iii) 7136985 = 7 + 3 + 9 + 5 = 24

and 1 + 6 + 8 = 15

Difference = 24-15 = 9

divisible by 11

#### Question 5. Examine the following numbers for divisibility by 6:

(i) 93573
(ii) 217944
(iii) 5034126
(iv) 901352
(v) 639210
(vi) 1790184

A number is divisible by 6 if it is divisible by 2 as well as by 3

(i) 93573 : It is not divisible by 6 because, it is not divisible by 2

(ii) 217944 : divisible by 6 if it is divisible by 2 as well as by 3

so the number is divisible by 2

Hence, given number is divisible by 6

(iii) 5034126 : divisible by 6 if it is divisible by 2 as well as by 3

(iv) 901352 : In this number is last digit is 2 , which is divisible by 2

now the sum of  number is – 9+0+1+3+5+2  = 20

20 is not divisible by 3

so the given number is not divisible by 6

(v) 639210 : In this number is last digit is 4 , which is divisible by 2

now the sum of  number is – 6+3+9+2+1+0  = 21

21 is divisible by 3

so the given number is divisible by 6

(vi) 1790184 : In this number is last digit is 4 , which is divisible by 2

now the sum of  number is – 1+7+9+0+1+8+4  = 30

30 is divisible by 3

so the given number is divisible by 6

#### Question 6. In each of the following replace ‘*’ by a digit so that the number formed is divisible by 9:

(i) 4710 * 82
(ii) 70 * 356722

(i) 4710 * 82

sum of the given number are =  4+7+1+0+8+2 = 22

the number is 22 which is divisible by 9 is 27.

smallest number = 27 – 22 = 5

(ii) 70 * 356722

sum of the given number are =  7+0+3+5+6+7+2+2 = 32

the number is 32 which is divisible by 9 is 36.

smallest number = 36 – 32 = 43

#### Question 7. In each of the following replace ‘*’ by (i) the smallest digit (ii) the greatest digit so that the number formed is divisible by 3:

(a) 4 * 672
(b) 4756 * 2

(a) 4 * 672

(i) the smallest digit

sum of the given number = 4+6+7+2 = 19

19 is not divisible by 3

so smallest digit is = 2

(i) the greatest digit

The greatest digit is 8

ex. 19 + 8 = 27

which is divisible by 3

(b) 4756 * 2

(i) the smallest digit

sum of the given number = 4+7+5+6 = 24

24 is divisible by 3

so smallest digit is = 0

(i) the greatest digit

The greatest digit is 9

ex. 24  + 9 = 33

which is divisible by 3

#### Question 8. In each of the following replace ‘*’ by a digit so that the number formed is divisible by 11:

(i) 8 * 9484
(ii) 9 * 53762

(i) 8 * 9484

Sun of the given digits from the right
= 4 + 4 + required digit
= 8 + required digit
Sum of the given digits  from the right = 8+9+8 = 25
Difference of sums = 25-(8+ required digit) = (17-required digit)
11 is the number smaller than 17, who gets divided by 11
so For the above difference to be divisible by 11 required digit = 6
Hence the required number is 869784

(ii) 9 * 53762
Sum of the given digits from the right = 2+7+ 5+9 = 23
Sum of the given digits from the right
= 6 + 3 + required number = 9
Difference of sums = 23 (9+ required number) = 14- required number
For the above difference to be divisible by 11 required digit- 3
= 14 – 3 = 11

11 is divisible by 11

Hence, the required number is 9353762

#### Question 9. In each of the following replace ‘*’ by (i) the smallest digit 00 the greatest digit so that the number formed is divisible by 6:

(a) 2 * 4706
(b) 5825 * 34

(a) 2* 4706

If the number is divisible by 6 then the number should also get divisible by 2 and 3
The last digit of 2* 4706 is 6,

so it is divisible by 2.

The sum of 2 * 4706
= 2+4+7 0+6 = 19(i) Smallest required number to be added in 19 is 2.
As 19+2=21    …….(ex. 21 is divisible by 3)
(ii) Greatest required number to be added in 19 is 8
As 19 +8 = 27    …… (ex.. 27 is divisible by 3)
(b) 5825*34
If the number is divisible by 6, then it should get divisible by 2 and 3.
The last number is 4,
so it is divisible by 2
The sum of 5825*34
= 5+8+2 +5+3+ 4 = 27
(i) The smallest number to be added in 27 is 0
27+0 = 27      ….. (27 is divided by 3)
(ii) Greatest number to be added in 27 is 9
ex. 27+9 = 36       ……. 36 is divided by 3

#### Question 10. Which of the following numbers are prime:

(i) 101
(ii) 251
(iii) 323
(iv) 397

(i) 101

We have, 101 = 1 x 101
101 has exactly two factors 1 and 101 itself.
so 101 is a prime number.

(ii) 251

We have, 251 = 1 x 251

251 has exactly two factors 1 and 251 itself.
so 251 is a prime number.

(iii) 323

We have, 323 = 1 x 323

= 17 x 19
Factors of 323 are 1, 17, 19, 323
323 has more than two factors.
so 323 is not a prime number.

(iv) 397

We have, 397 =  1 x 397
397 has exactly two factors 1 and 397 itself.

so 397 is a prime number.

#### Question 11. Determine if 372645 is divisible by 45.

Determine if 25110 is divisible by 45, we test it for divisible by 5 and 9 both.
Divisibility of 372645 by 5

so Number in the unit’s place of 372645 = 5

372645 is divisible by 5

Divisibility of 372645 by 9

Sum of the digits of the number 372645 = 3+7+2 +6+4+5 = 27    ……..(27 is divisible by 9)
so 372645 is divisible by 9
As 372645 is divisible by 5 and 9 both and 5 and 9 are co-prime numbers,

so 372645 is
divisible by 5 x 9= 45

#### Question 12. A number is divisible by 12. By what other numbers will that number be divisible?

The number is divisible by 12 ,

should also get divisible by all its factors.

So, the number by which given number is divisible are : 1,2,3,4,6.

#### Question 13. A number is divisible by both 3 and 8. By which other numbers will that number be always divisible?

A natural number, say n, be divisible by both 3 and 8.

As 3 and 8 are co-prime numbers using property n is divisible by 3 x 8

(i.e. 24).
Thus, the given number is always divisible by 24.
So, the given number should get divided by all the factors of 24.

Therefore, the other number by which the given numbers is always divisible are: 1, 2, 4, 6, 12, 24

#### Question 14. State whether the following statements are true (T) or false (F):

(i) If a number is divisible by 4, it must be divisible by 8.
(ii) If a number is divisible by 3, it must be divisible by 9.
(iii) If a number is divisible by 9, it must be divisible by 3.
(iv) If a number is divisible by 9 and 10 both, it must be divisible by 90.
(v) If a number divides the sum of two numbers, then it must divide the two numbers separately. .
(vi) If a number is divisible by 3 and 8 both, it must be divisible by 12.
(vii) If a number is divisible by 6 and 15 both, it must be divisible by 90.

(i) False
(ii) False
(iii) Ture
(iv) Ture
(v) False
(vi) Ture
(vii)  False

—  : End of ML Aggarwal Playing With Numbers Exe-4.2 Class 6 ICSE Maths Solutions :–

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