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ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.3

July 11, 2019 by Prasanna

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 5 Playing with Numbers Ex 5.3

Question 1.
Which of the following numbers are divisible by 5 or by 10:
(i) 87035
(ii) 75060
(iii) 9685
(iv) 10730
Solution:
A number having its unit digit 5 or 0, is divisible by 5.
A number having its unit digit 0, is divisible by 10.
So, 87035, 75060, 9685, 10730 are all divisible by 5
but number 75060 and 10730 are divisible by 10.

Question 2.
Which of the following numbers are divisible by 2, 4 or 8:
(i) 67894
(ii) 5673244
(iii) 9685048
(iv) 6533142
(v) 75379
Solution:
A number having its unit digit 2, 4, 6, 8 or 0 is divisible by 2,
a number is divisible by 4, if the number formed by the
last two digits is divisible by 4 and a number is divisible by 8,
of the number formed by the last three digits is divisible by 8. So,
Number 67894, 5673244, 9685048, 6533142 are divisible by 2.
Numbers, 5673244, 9685048 are divisible by 4
and numbers 9685048 is divisible by 8.

Question 3.
Which of the following numbers are divisible by 3 or 9:
(i) 45639
(ii) 301248
(iii) 567081
(iv) 345903
(v) 345046
Solution:
A number is divisible by 3 if the sum of its digits is divisible by 3,
a number is divisible by 9 if the sum of its digits is divisible by 9. So,
45639, 301248, 567081, 345903 are divisible by 3.
49639, 301248, 467081 are divisible by 9.

Question 4.
Which of the following numbers are divisible by 11:
(i) 10835
(ii) 380237
(iii) 504670
(iv) 28248
Solution:
A number is divisible by 11 if the difference of the sum of digits
at the odd places and sum of the digits at even places is zero or divisible by 11. So,
10835, 380237, 28248 are divisible by 11.

Question 5.
Which of the following numbers are divisible by 6:
(i)15414
(ii) 213888
(iii) 469876
Solution:
A number is divisible by 6 if it is divisible by 2 as well as by 3,
so 15414 and 213888 are divisible by 6.

Question 6.
Which of the following numbers are divisible by 7:
(i) 46f8894875
(ii) 3794856
(iii) 39823
Solution:
A number is divisible by 7 if the difference of the sum of digits in
alternate blocks of three digits from the right to the left is divisible by 7,
so 4618894875 and 39823 are divisible by 7.

Question 7.
(i) If 34x is a multiple of 3, where x is a digit, what is the value of x?
(ii) If 74×5284 is a multiple of 3, where x is a digit, find the value(s) of x.
Solution:
(i) 34x is a multiple of 3
lf 3 + 4 + x = 7 + x is divisible by 3
Then x + 7 = 9 ⇒ x = 9 – 7 = 2
∴ x = 2, 5, 8

(ii) 74×5284 is divisible by 3
∴ 7 + 4 + x + 5 + 2 + 8 + 4 is divisible by 3
⇒ 30 + x is divisible by 3
x = 0, 3, 6, 9

Question 8.
If 42z3 is a multiple of 9, where z is a digit, what is the value of z?
Solution:
42z3 is a multiple of 9
∴ 4 + 2 + z + 3 is divisible by 9
⇒ 9 + z is divisible by 9
Either 9 + z = 9, or 9 + z = 0
∴ z = 9 + 9 = 18, Or z = 9 – 9 = 0
∴ z = 0, 9

Question 9.
In each of the following replace * by a digit so that the number formed is divisible by 9:
(i) 49 * 2207
(ii) 5938 * 623
Solution:
(i) 49×2207 is divisible by 9
If 4 + 9 + x + 2 + 2 + 0 + 7 is divisible by 9
⇒ 24 + x is divisible by 9 ⇒ 24 + x = 27
⇒ x = 27 – 24 is divisible by 9
∴ x = 3

(ii) 5938×623 is divisible by 9
If 5 + 9 + 3 + 8 + x + 6 + 2 + 3 is divisible by 9
If 36 + x is divisible by 9
⇒ 36 + x = 36 or 45
∴ x = 36 – 36 = 0 or x = 45 – 36 = 9
∴ x = 0, 9

Question 10.
In each of the following replace * by a digit so that the number formed is divisible by 6:
(i) 97 * 542
(ii) 709 * 94
Solution:
(i) 97 * 542 is divisible by 6
∴ It is divisible by 2 and 3
∵ Unit digit is 2
∴ It is divisible by 2
∵ It is divisible by 3
∴ Sum of its digits 9 + 7 + 5 + 4 + 2 = 27
which is divisible by 3 27 + * = 27, or 30, 33, 36
∴ The blank * will be 0 or 3 or 6 or 9

(ii) 709 * 94
∴ It is divisible by 6
⇒ It is divisible by 2 and 3
But its unit digit is 4
∴ It is divisible by 2
Now sum of its digits = 7 + 0 + 9 + 9 + 4 + * = 29 + *
∵ It is divisible by 3
29 + * = 30, 33, 36,
∴ Blank (*) = 1 or 4 or 7

Question 11.
In each of the following replace * by a digit so that the number formed is divisible by 11:
(i) 64*2456
(ii) 86*6194
Solution:
(i) 64 * 2456 is divisible by 11
The difference between the sum of digits of odd places
– Sum of digits of even place is divisible by 11 it is zero.
6 + 4 + * + 6 – 5 + 2 + 4 is divisible by 11
⇒ 16 + * – 11 is divisible by 11
⇒ 5 + x is divisible by 11
∴ * = 6

(ii) 86 * 6194
∵ It is divisible by 11
The difference between the sum of digits of odd places and
sum of digits of even places is divisible by 11 or it is zero.
Now, 4 + 1 + * + 8 = 13 + *
9 + 6 + 6 = 21
∴ 21 – (13 + *) is divisible by 11
⇒ 21 – 13 – * is divisible by 11
⇒ 8 – * is divisible by 11
∴ * is 8

ML Aggarwal Class 8 Solutions for ICSE Maths

Filed Under: ICSE

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