ML Aggarwal CBSE Solutions Class 8 Math Fifth Chapter Playing with Numbers Exercise 5.3

ML Aggarwal CBSE Solutions Class 8 Math 5th Chapter Playing with Numbers Exercise 5.3

(1) (i) 87035

=> ‘87035’ its last digit ‘5’

Therefore, it is divided by ‘5’

(ii) 75060

=> Its last digit ‘0’ it is divided ‘10’, ‘5’

(iii) 9680

Its last digit ‘0’ it is divided by ‘5’ & ‘10’

(iv) 10730

=> Its last digit ‘0’ it is divided ‘10’ & ‘5’

(2) (i) 67894

=> It last digit ‘4’ it is divisible by ‘2’

(ii) 5673244

=> It last digit ‘4’ & last two digits ‘44’ it is divisible by ‘2’ & ‘4’

(iii) 9685048

=> it is divisible by ‘2’ & ‘4’

(iv) 6533142

=> it is divisible by ‘2’

(v) 75379

=> Its last digit ‘9’ it is not divisible by ‘2’, ‘4’, ‘8’

(3) (i) 45639

=> 4 + 5 + 6 + 3 + 9 = 27

Therefore, it is divisible by ‘3’ & ‘9’

(ii) 301248

=> 3 + 0 + 1 + 0 + 4 + 8 = 10

Therefore, It is divisible by ‘3’ & ‘9’

(iii) 567081

=> 5 + 6 + 7 + 0 + 8 + 1 = 27

Therefore, it is divisible by ‘3’ & ‘9’

(iv) 345908

=> 3 + 4 + 5 + 9 + 0 + 8 = 29

Therefore, It is divisible.

(v) 345046

=> 3 + 4 + 5 + 0 + 4 + 5 6 = 22

It is not divisible.

 (4) (i) 10835

=> Sum of digits of odd places = 1 + 3 + 5 = 9

Sum of digits of even places = 8

Therefore, difference = 9 – 8 = 1

Therefore, it is not divisible by ‘11’

(ii) 380237

=> Sum of digits of odd places = 3 + 3 + 7 = 13

Sum of digits of even places = 8 + 2 = 10

Therefore, difference = 13 – 10 = 3

Therefore, it is not divisible by ‘11’

(iii) 504670

=> Sum of digits of odd places = 5 + 7 = 12

Sum of digits of even places = 4 + 6 = 10

Therefore, their difference = 12 – 10 = 2

Therefore, it is not divisible by ‘11’

(iv) 28248

=> Sum of digits of odd places = 0

Sum of digits of even places = 2 + 8 + 2 + 4 + 8 = 22

Their difference = 22 – 0 = 22

Therefore, it is divisible by ‘11’

(5) (i) 15414

=> its last digit ‘4’ it is divisible by ‘2’

Also, 1 + 5 + 4 + 1 + 4 = 15

It is divisible by ‘3’

Therefore, ‘15414’ is divisible by ‘6’

(ii) 213888

=> Its last digit ‘6’

It is divisible by ‘2’

Also, 4 + 6 + 9 + 8 + 7 + 6 = 40

It is not divisible by ‘3’

Therefore, ‘469876’ is not divisible by ‘6’

(6) Sum of digit = 3 + 4 + x = 7 + x

‘7 + x’ is possible when,

7 + x = 7, 14, 21 …..

Or, x = 0, 7, 14, 21

Therefore, since x is a digit

Therefore, x = 0, 7

(7) Sum of digit = 4 + 2 + Z + 3 = 9 + Z

‘9 + z’ is divisible by ‘9’

This is possible when—

9 + z = 9, 18, 27 …

Or, z = 0, 9, 18, 27 …..

Therefore, since, z is digits

Because, z = 0, 9

(8) (i) 49 * 2207

=> Sum of digits = 7 + 0 + 2 + 2 + 9 + 4

= 24

If, we add ‘3’

It becomes = 24 + 3 = 27

It is divisible by ‘9’

Because, ‘*’ is to be replaced by ‘3’

Therefore, the number formed = 4932207

Which is divided by ‘9’

(ii) 5938*623

=> Sum of digits = 3 + 2 + 6 + 8 + 3 + 9 + 5 = 36

Therefore, ‘*’ is to be replaced by ‘0’

Therefore, the number formed = 59380623

Which is divisible by ‘9’

(9) (i) 97 * 542

=> sum of digit = 2 + 4 + 5 + 7 + 9 = 27

We should be add 3.

It become ‘30’

Therefore, * is to be replaced by ‘3’

Therefore, 97 * 542 = 973542

Therefore, the number formed = 973542

Which is divisible by ‘6’.

(ii) 709 * 94

=> Sum of digit = 4 + 9 + 9 + 0 + 7 = 29

We should be add 1

It become ‘30’

Therefore, ‘*’ is to be replaced by ‘1’

Therefore, the number formed = 709194

Which is divide by ‘6’

(10) (i) 64 * 2456

=> sum of digit = 6 + 5 + 4 + 2 + 4 + 6 = 27

We need to add 6.

It became ‘30’

Therefore, ‘*’ is to be replaced by ‘6’

Since the number formed = 6462456

Which is divisible by 11

(ii) 86 * 6194

=> Sum, of digit = 4 + 9 + 1 + 6 + 6 + 8 = 34

We need to add ‘9’

It become 44

Therefore, ‘*’ is to be replaced by 44.

Since the number formed = 8696194

Which is divisible by ‘11’

Updated: March 25, 2021 — 6:19 pm

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  1. You are genius bro

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