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’
You are genius bro